Full text of "A Text-book of Euclid's Elements for the Use of Schools"

See other formats

Google

This is a digital copy of a book that was preserved for generations on library shelves before it was carefully scanned by Google as part of a project

to make the world's books discoverable online.

It has survived long enough for the copyright to expire and the book to enter the public domain. A public domain book is one that was never subject

to copyright or whose legal copyright term has expired. Whether a book is in the public domain may vary country to country. Public domain books

are our gateways to the past, representing a wealth of history, culture and knowledge that's often difficult to discover.

Marks, notations and other maiginalia present in the original volume will appear in this file - a reminder of this book's long journey from the

publisher to a library and finally to you.

Usage guidelines

Google is proud to partner with libraries to digitize public domain materials and make them widely accessible. Public domain books belong to the
public and we are merely their custodians. Nevertheless, this work is expensive, so in order to keep providing tliis resource, we liave taken steps to
prevent abuse by commercial parties, including placing technical restrictions on automated querying.
We also ask that you:

+ Make non-commercial use of the files We designed Google Book Search for use by individuals, and we request that you use these files for
personal, non-commercial purposes.

+ Refrain fivm automated querying Do not send automated queries of any sort to Google's system: If you are conducting research on machine
translation, optical character recognition or other areas where access to a large amount of text is helpful, please contact us. We encourage the
use of public domain materials for these purposes and may be able to help.

+ Maintain attributionTht GoogXt "watermark" you see on each file is essential for in forming people about this project and helping them find
additional materials through Google Book Search. Please do not remove it.

+ Keep it legal Whatever your use, remember that you are responsible for ensuring that what you are doing is legal. Do not assume that just
because we believe a book is in the public domain for users in the United States, that the work is also in the public domain for users in other
countries. Whether a book is still in copyright varies from country to country, and we can't offer guidance on whether any specific use of
any specific book is allowed. Please do not assume that a book's appearance in Google Book Search means it can be used in any manner
anywhere in the world. Copyright infringement liabili^ can be quite severe.

About Google Book Search

Google's mission is to organize the world's information and to make it universally accessible and useful. Google Book Search helps readers
discover the world's books while helping authors and publishers reach new audiences. You can search through the full text of this book on the web

at |http: //books .google .com/I

ia^ 3^ /fO if-.-

:'f

By H. S. hall.

Algebraical Examples Supplementary to Hall and Knight's
Algebra for Beginners and Elementary Algebra. (Chaps.

I.-XXVIl.)- With or without Answers. Globe 8vo. 2s.

Sbort Introduction to Graphical Algebra Revised Edition. Gi.

8vo. Is. [Key, in the press.

By H. S. HALL and S. R. KNIGHT.
Elementary Algebra for Schools Ss. 6d with Answers. Globe

8vo. 4s. 6d. Key. 88. 6d.

Answers to Examples in Elementary Algebra Fcap. 8vo.

Sewed. Is.

Algebraical Exercises and Examination Papers, with or with-
out Answers. Globe 8vo. 28. 6d.

Higher Algebra A Sequel to Elementary Algebra for Schools. Or.
8vo. 7s. 6d. Key. 10s. 6d.

Algebra for Beginners Globe 8vo. 2s. with Answers. 2s. 6d.

Arithmetical Exercises and Examination Papers with or

without Answers. Globe 8vo. 28. Gd.

Elementary Trigonometry Globe 8vo. 4s. 6d. Key, 8s. 6d.

By H. S. HALL and F. H. Sl'EVENS.

A Text-Book of Euclid's Elements. Containing Books I. -VI., XL,
and XII. Props. 1 and 3. Globe 8vo. 4s. 6d. Or separately

Book I. Is.

Books I. and II. Is. 6d.

Books II. and III. 2s.

Books I. -III., 23. 6d. ; sewed, 28.

Books I. -I v., 3s. ; sewed, 23. 6d.

Books III. and IV. 28.
Books III. -VI. 3s.
Books IV.. VI. 28. fid.
Books v., VI., XL, and XIL, 1
Book XL Is. [and 3. 28. 6d.

A Key to the Exercises and Examples contained in a Text-

BOOK of Euclid's Elements Books I.-VI. and XL 8s. Gd.
Books I. -IV. Or. 8vo. 6s. 6d. Books Vl. and XI. 3s. 6d.

A School Geometry, based on the recommendations of the Mathe-
matical Association, and on the recent report of the Cambridge
Syndicate on Geometry. GL 8vo.

Parts I. to IV. 8s.
Part V. Is. 6d.
Parts I to V. 4s. 6d.
Parts IV. and V. 2s.
Part VI. 28.

Parts I and II. Is. 6d.
Part IIL Is.

Parts L, IL, and IIL 2s. 6d.
Parts III. and IV. Is. 6d.
Part IV. Sewed, 6d.

An Elementary Course of Mathematics Globe 8vo. 2s. cd.

By F. H. STEVENS.

Elementary Mensuration Globe 8vo. 3s. 6d.
Mensuration for Beginners Globe 8vo. is. 6d.

By H. S. HALL and R. J. WOOD.
Algebra for Elementary Schools Globe 8vo. Parts i., ii., and

IIL, 6d. each. Cloth, 8d. each. Answers to each Part, 4d. each.
Sewed.

LONDON: MACMILLAN AND CO., LIMITED.

A TEXT-BOOK OF EUCLID'S ELEMENTS

A TEXT-BOOK

EUCLID'S ELEMENTS

FOR THE USE OF SCHOOLS

BOOKS L-VI. AND XI.

H. S. HALL, M.A.

FORMERLT SCHOLAR OF CHRIST'S COLLEGE, CAMBRIDGE;

AND

F. H. STEVENS, M.A.

FORMERLY SCHOLAR OF QUEEN'S COLLEGE, OXFORD.

NEW EDITION

MACMILLAN AND CO., Limited

NEW YORK: THE MACMILLAN COMPANY

1904

All rights rcaerocd.

First Edition, 1888. Second Edition (Book XI. added), 1889.

Beprintod 1890, 1891, 1892, 1898, 1894, 1895 (twice), 1896, 1897, 1898, 1899.

New Edition 1900. Keprinted 1901, 1902, 1903, 1904.

OLASQOW: PRINTED AT THE UNIVERSITY PRESS
BY ROBERT MACLEHOSE AND CO. LTD.

EXTRACT FROM THE PREFACE TO THE

FIRST EDITION.

This volume contains the first Six Books and part of the
Eleventh Book of Euclid's Elements, together with additional
Theorems and Examples, giving the most important elementary
developments of Euclidean Geometry.

The text has been carefully revised, and special attention
given to those points which experience has shewn to present
difficulties to beginners.

In the course of this revision the enunciations have been
altered as little as possible : and very few departures have
been made from Euclid's proofs ; in each case changes have
been adopted only where the old text has been generally found
a cause of difficulty ; and such changes are for the most part
in favour of well-recognised alternatives.

In Book I., for example, the ambiguity has been removed
from the Enunciations of Propositions 18 and 19, and the fact
that Propositions 8 and 26 establish the complete equality of
the two triangles considered has been strongly urged : thus
the redundant step has been removed from Proposition 34.

In Book II. Simson's ariangement of Proposition 13 has been
abandoned for a well-known equivalent.

In Book III. Propositions 35 and 36 have been treated
generally, and it has not been thought necessary to do more
tlian call attention in a note to the special cases.

These are the chief deviations from the ordinary text as
regards method and arrangement of proof ; they are points
familiar as difficulties to most teachers, and to name them
indicates sufficiently, without further enumeration, the general
principles which have guided our revision.

First Edition, 1888. Second Edition (Book XI. added), 1889.

Beprintod 1890, 1891, 1892, 1898, 1894, 1895 (twice), 1896, 1897, 1898, 1899.

New Edition 1900. Keprinted 1901, 1902, 1903, 1904.

GLASGOW: PRINTED AT THE UNIVERSITY PRESS
BY ROBERT MACLEUOSE AND CO. LTD.

EXTRACT FROM THE PREFACE TO THE

FIRST EDITION.

The text has been carefully revised, and special attention
given to those points which experience has shewn to present
difl&culties to beginners.

In Book II. Simson's an angement of Proposition 13 has been
abandoned for a well-known equivalent.

In Book III. Propositions 35 and 36 have been treated
generally, and it has not been thought necessary to do more
tlian call attention in a note to the special cases.

VI PREFACE.

A few alternative proofs of difficult propositions are given for
the convenience of those teachers who care to use them.

One purpose of the book is gradually to familiarise the student
with the use of legitimate symbols and abbreviations; for a
geometrical argument may thus be thrown into a form which
is not only more readily seized by an advanced reader, but is
useful as a guide to the way in which Euclid's propositions may
be handled in written work. On the other hand, we think it
very desirable to defer the introduction of symbols until the
beginner has learnt that they can only be properly used in
Pure Geometry as abbreviations for verbal argument : and we
hope thus to prevent the slovenly and inaccurate habits which
are very apt to arise from their employment before this principle
is fully recognised.

Accordingly in Book I. we have used no contractions or
symbols of any kind, though we have introduced verbal altera-
tions into the text wherever it appeared that conciseness or
clearness would be gained.

In Book JI. abbreviated forms of constantly recurring words
are used, and the phrases therefore and is eqxwtl to are replaced
by the usual symbols.

In the Third and following Books, and in additional matter
throughout the whole, we have employed all such signs and
abbreviations as we believe to add to the clearness of the
reasoning, care being taken that the symbols chosen are com-
patible with a rigorous geometrical method, and are recognised
by the majority of teachers.

If this arrangement should be thought fanciful or wanting in
uniformity, we may plead that it is the outcome of long experi-
ence in the use of various text-books. For some years, for
example, we were accustomed to teach from a symbolical text,
but in consequence of the frequent misconceptions and inac-
curacies which too great brevity was found to generate among
beginners, we were compelled to return to one of the older
and unabbreviated editions. The gain to our younger boys
was immediate and unmistakeable ; but the change has not

PREFACE. VU

been unattended with disadvantage to more advanced students,
who on reachuig the Third or Fourth Book may not only be
safely trusted with a carefully chosen system of abbreviations,
but are certainly retarded by the monotonous and lengthy
formalities of the old text.

It must be understood that our use of symbols, and the
removal of imnecessary verbiage and repetition, by no means
implies a desire to secure brevity at all hazards. On the con-
trary, nothing appears to us more mischievous than an abridge-
ment which is attained by omitting steps, or condensing two
or more steps into one. Such uses spring from the pressure of
examinations ; but an examination is not, or ought not to be, a
mere race ; and while we wish to indicate generally in the later
books how a geometrical argument may be abbreviated for the
purposes of written work, we have not attempted to reduce the
propositions to the barest skeleton which a lenient Examiner
may be supposed to accept. Indeed it does not follow that the
form most suitable for the page of a text-book is also best adapted
to examination purposes ; for the object to be attained in each
case is entirely different. The text-book should present the
argument in the clearest possible manner to the mind of a
reader to whom it is new : the written proposition need only
convey to the Examiner the assurance that the proposition has
been thoroughly grasped and remembered by the pupil.

From first to last we have kept in mind the undoubted fact
that a very small proportion of those who study Elementary
Geometry, and study it with profit, are destined to become
mathematicians in any special sense ; and that, to a large
majority of students, Euclid is intended to serve not so much
as a first lesson in mathematical reasoning, as the first, and
sometimes the only, model of formal argument presented in an
elementary education.

This consideration has determined not only the full treatment
of the earlier Books, but the retention of the formal, if some-
what cumbrous, methods of Euclid in many places where proofs
of greater brevity and mathematical elegance are a.\«A\aXi\fe»

VUl PREFACE.

We hope that the additional matter introduced into the book
will provide sufficient exercise for pupils whose study of Euclid
is preliminary to a mathematical education.

The questions distributed through the text follow very easily
from the propositions to which they are attached, and we think
that teachers are likely to find in them all that is needed for an
average pupil reading the subject for the first time.

The Theorems and Examples at the end of each Book contain
questions of a slightly more difficult type : they have been very
carefully classified and arranged, and brought into close connec-
tion with typical examples worked out either partially or in
full ; and it is hoped that this section of the book, on which
much thought has been expended, will do something towards
removing that extreme want of freedom in solving deductions
that is so commonly found even among students who have a
good knowledge of the text of Euclid.

To Volumes containing only Books I.-III., or Books I.-IV. an
Appendix is added, giving an elementary account of the
properties of Pole and Polar, and Eadical Axis. In the com-
plete book these subjects, together with a short account of
Harmonic Section, Centres of Similitude, and Transversals,
appear as Theorems and Examples on Book VL

• •••••••

Throughout the book we have italicised those deductions on
which we desired to lay special stress as being in themselves
important geometrical results ; this arrangement we think will
be useful to teachers who have little time to devote to riders, or
who wish to sketch out a suitable course for revision.

H. S. HALL.
F. H. STEVENS.

Clifton, December, 1886.

PEEFATORY NOTE TO THE NEW EDITION.

Is the present edition the text has received further revision,
and the notes have been for the most part re-written, with a
view to greater clearness and simplicity.

References to the Definitions being frequent in the text of
Book I., the convenience of a standard order has been pointed
out to us by many elementary teachers. We have therefore
thought it advisable to re-number the Definitions in accordance
with Simson's edition. This has involved the insertion of
certain definitions hitherto omitted as of slight importance :
such insertions have now been printed in subordinate type.

A few typographical improvements have been introduced :
notably the italicising of Particular Enunciations, Some changes
in pagination have also been effected for the purpose of pre-
senting the whole of a proposition at one view, or of bringing
notes and exercises into closer connection with the text to
which they refer. Further, the symbols " /. '' for therefore,
and " = " for is equal to are now introduced from the 35th
Proposition of Book I.

Groups of Test Questions for Revision have been inserted
at various stages. These may be useful to beginners, and
snggestive to teachers in framing examination papers, which
so often consist of mere monotonous lists of propositions and
examples.

One important change has been made. The algebraical
treatment of the subject-matter of Book Y. has been entirely
separated from the stricter general treatment, so as to present
in the simplest form such Definitions and Theorems of Pro-
portion as are necessary before entering upon Book VI. This
Introduction will be found immediately preceding Book VL
in a chapter called The Elementary Principles of Proportion,

s. s. s.

F. H. S.
Febrvary, 1900.

CONTENTS

BOOK I.

PAGE

Definitions, Postulates, Axioms 1

Section I. Propositions 1-26 12

Section II. Parallels and Parallelograms.

Propositions 27-34 ..... 66

Section III. The Areas of Parallelograms and Triangles.

Propositions 35-48 72

Theorems and Examples on Book I.

Analysis, Synthesis 95

I. On the Identical Equality of Triangles . . 98

n. On Inequalities 101

m. On Parallels 103

IV. On Parallelograms ....... 104

V. Miscellaneous Theorems and Examples . . 108

VI. On the Concurrence of Straight Lines in a

Triangle 110

Vil. On the Construction of Triangles with given

Parts 115

Vm. On Areas 117

IX. On Loci 122

X. On the Intersection of Loci "V^

Xn CONTENTS,

BOOK 11.

PAGE

Definitions, etc c . . 128

Propositions 1-14 130

Theorems and Examples on Book II 158

BOOK III.
Definitions, etc 163

Propositions 1-37 167

Note on the Method of Limits as Applied to Tangency 231

Theorems and Examples on Boole III.

I. On the Centre and Chords of a Circle . . 233

II. On the Tangent and the Contact of Circles.

The Common Tangent to Two Circles, Problems on

Tangency, Orthogonal Circles .... 235

III. On Angles in Segments, and Angles at the Cen-
tres AND Circumferences of Circles.
The Orthocentre of a Triangle, and Properties of

the Pedal Triangle, Loci, Simson's Line . . 240

rV. On the Circle in Connection with Rectangles.

Further Problems on Tangency .... 251

V. On Maxima and Minima 257

VI. Harder Miscellaneous Examples .... 264

BOOK IV.
Definitions, 268

Propositions 1-16 269

Note on Regular Polygons 294

Theorems and Examples or Booh IV.

I. On the Triangle and its Circles.

Circumscribed, Inscribed, and Escribed Circles,

The Nine-Points Circle . . .297

JI. Miscellaneous Examples 303

CONTENTS. XUl
BOOK V.

PAGE

I>EFINinONS, . . . . , 305

Propositions 1-16, 309

EUmmtary Principles of Proportion,

Introduction to Book VI 317

BOOK VL

Definitions 325

Propositions 1-D. 326

Theorems and Examples on Book VI.

L On Harmonic Section 384

n. On Centres of Similarity and Similitude » . 388

in. On Pole and Polar . , 390

rv. On the Radical Axis of Two or More Circles . 396

V. On Transversals 399

VL Miscellaneous Examples on Book VI. . , . 403

BOOK XI.

Definitions 409

Propositions 1-21 419

Exercises on Book XI 444

Theorems and Examples on Book XI. ... 446

BOOK XII.

« «

The First Proposition of Book XII. xoill he found at p. .S64,
worked out cw an Example on Proposition 20 of Book VI. ^
of which it is a development. Prop. 3 of Book XII. is
briefly treated as a Corollary to Prop. 1.

•><r-

J)EF/ /y I TIO /VS divided -into Sect^o^S of 5.

EUCLID'S ELEMENTS.

BOOK 1.

Definitions.'

1. A point is that which has position, but no magnitude.

2. A line is that which has length without breadth.

3. The extremities of a line are points, and the intersection of
two lines is a point. /

4. A straight line is that which lies evenly between its
extreme points.

Any portion cut off from a straight line is called a segment of it.

5. A surfskce (or superficies) is that which has length
and breadth, but no thickness.

6. The boundaries of a surface are lines.

7. A plane surface is one in which any two points being (^
taken, the straight line between them lies wholly in that
surface.

A plane surface is frequently referred to simply as a plane.

Note. Euclid regards a point merely as a mark of position, and
he therefore attaches to it no idea of size and shape.

Similarly he considers that the properties of a line arise only
from its length and position, without reference to that minute
breadth which every line must really have if actvxjdly dravm, even
though the most perfect instruments are used.

The definition of a surface ia to be understood m e. «vtcSl^x '^^'^^

EUCLID'S ELEMENTS.

f^-

8. A plane angle is the inclination of two
lines to one another, which meet together, but
are not in the same direction.

[Definition 8 is not required in Euclid's Geometry, the only angles
employed by him being those formed by straight lines. See Def . 9. ]

9. A plane rectilineal angle is the inclina-
tion of two straight lines to one another, which
meet together, but are not in the same straight
line. —

The point at which the straight lines meet is called the yertez of
the angle, and the straiglit lines themselves the arms of the angle.

Note. When there are several angles at one
point, each is expressed by three letters, of which
the letter that refers to the vertex is put between
the other two. Thus the angle contained by the
straight lines OA, OB is named the angle AOB
or BOA ; and the angle contained by OA, OC is
named the angle AOC or COA. But if there is
only one angle at a point, it may be expressed by
a single letter, as the angle at O.

Of the two straight lines OB, OC shewn in the
adjoining diagram. Me recognize that OC is more
inclined than OB to the straight line OA : this we
express by saying that the angle AOC is greater
than the angle AOB. Thus an angle must be
regarded as having magnitude. q

It must be carefully observed that the size of an angle in no way
depends on the length of its arms, but only on their inclination to
one another.

The angle AOC is the svm of the angles AOB and BOC ; and
AOB is the difference of the angles AOC and BOC.

[Another view of an angle is recognized in many branches of
mathematics ; and though not employed by Euclid, it is here given
because it furnishes more clearly than any other a conception of
what is meant by the magnitude of an angle.

Suppose that the straight line OP in the diagram
is capable of revolution about the point O, like
the hand of a watch, but in the opposite direction ;
and suppose that in this way it has passed suc-
cessively from the position OA to the positions
occupied by OB and OC. Such a line must have
undergone rnore turning in passing from OA to

DEFINITIONS.

OC, than in passing from OA to OB ; and consequently the angle
AOC is said to be greater than the angle AOB.]

Angles which lie on either side of a common
arm are called adjacent angles.

For example, when one straight line OC

is drawn from a point in another straight line

AB, the angles COA, COB are adjacent. £

When two straight lines, such as AB, CD,
cross one another at E, the two angles CEA,
BED are said to be yertically opposite. The
two angles CEB, AED are also vertically oppo-
site to one another.

10. When a straight line standing on
another straight line makes the adjacent
angles equal to one another, each of the
angles is called a right angle ; and the
straight line which stands on the other is
called a perpendicular to it.

11. An obtuse angle is an angle which
is greater than a right angle.

12. An acute angle is an angle which is
less than a right angle.

fin the adjoining figure the straight line
OB may be supposed to have arrived at its

E resent position, from the position occupied
y OA, by revolution about the point O in
either of the two directions indicated by the
arrows : thus two straight lines drawn from
a point may be considered as fonning two
angles (marked (i) and (ii) in the figure), of
which the greater (ii) is said to be reflex.

Tf the arms OA, OB are in the same
straight line, the angle formed by them q-
on either side is called a straight angle.]

z::i

4 Euclid's elements.

13. A term or boundary is the extremity of anything.

14. Any portion of a plane surface
bounded by one or more lines is called a
plane figure.

The sum of the bounding lines is called the perimeter of the figure.
Two figures are said to be equal in area when they enclose equal
portions of a plane surface.

15. A circle is a plane figure contained
by one line, which is called the circum-
ference, and is such that all straight lines
drawn from a certain point within the
figure to the circumference are equal to
one another; this point is called the
centre of the circle.

16. A radius of a circle is a straight line drawn from
the centre to the circumference.

17. A diameter of a circle is a straight line drawn
through the centre, and terminated both ways by the
circumference.

18. A semicircle is the figure
bounded by a diameter of a circle and
the part of the circumference cut off
by the diameter.

19. A segment of a circle is the figure
bounded by a straight line and the part
of the circumference which it cuts off.

20. Rectilineal figures are those which are bounded
by straight lines.

21. A triangle is a plane figure bounded
by three straight lines.

Any one of the angular points of a triangle may be regarded ae
its yertex ; and the opposite side is then called the base.

DEFINITIONS.

22. A quadrilateral is a plane figure
bounded by fmr straight lines.

The straight line which joins opposite angular
points in a quadrilateral is called a diia^onal.

23. A polygon is a plane figure
bounded by more than four straight lines.

Triangles.

24. An equilateral triangle is a triangle
whose three sides are equal.

25. An isosceles triangle is a triangle two
of whose sides are equal.

26. A scalene triangle is a triangle which
has three unequal sides.

27. A right-angled triangle is a triangle
which has a right angle.

The side opposite to the right angle in a right-angled triangle is
called the hsrpotenuse.

28. An obtuse-angled triangle is a
triangle which has an obtuse angle.

29. An acute-angled triangle is a triangle
which has three, acute angles.

[It will be seen hereafter (Book I. Proposition 17) that every
triangle must have at least two acute angles,]

Euclid's elements.

Quadrilaterals.

30. A square is a four-sided figure which
has all its sides equal and all its angles right
angles.

[It may be shewn that if a quadrilateral has all its
sides equal and one an^le a right angle, then cUl its
angles will be right angles.]

31. An oblong is a four-sided figure which has all its angles
right angles, but not all its sides equal.

32. A rhombus is a four-sided figure
which has all its sides equal, but its
angles are not right angles.

33. A rhomboid is a four-sided figure which has its opposite
sides equal to one another, but all its sides are not equal nor its
angles right angles.

34. All other four-sided figures are called trapeziums.

It is usual now to restrict the term trapezium to
a quadrilateral which has two of its sides parallel,
[See Def. 36.]

35. Paxallel straight lines are such as,
being in the same plane, do not meet, how-
ever far they are produced in either direc-
tion.

36. A Parallelogram is a four-sided
figure which has its opposite sides
parallel.

37. A rectangle is a parallelogram
which has one of its angles a right angle.

THE POSTULATES.

The Postulates.

Let it be granted,

1. Thai a straight line may he drawn from any one point to
any other point,

2. That a finite, that is to say a terminated^ straight line
may be produced to any length in that straight line,

3. TTiat a circle may be descnhed from any centre, at any
distance from that centre, that is, with a radius equal to any finite
straight line drawn from the centre.

Notes on the Postulates.

1. In order to draw the diagrams required in Euclid's Geometry
certain instruments are necessary. These are

(i) A ruler with which to draw straight lines,
(ii) A pair of compaasea with which to draw circles.

In the Postulates, or requests, Euclid claims the use of these
instruments, and assumes that they suffice for the purposes men-
tioned above.

2. It is important to notice that the Postulates include no means
of direct measurement : hence the straight ruler is not supposed to
be graducUed ; and the compasses are not to be employed for trann-
ftrring distances from one part of a diagram to another.

3. When we draw a straight line from the point A to the point
B, we are said to join AB.

To produce a straight line means to prolong or lengthen it.

The expression to describe is used in Geometry in the sense of
to draw.

On the Axioms.

The science of Geometry is based upon certain simple
statements, the truth of which is so evident that they are
accepted without proof.

These self-evident truths, called by Euclid Common
Notions, are known as the Axioms.

8 BtJCLID*S ELEMENTS,

General Axioms.

1. Things which are equal to the same thing are equal to one
another,

2. If equals he added to equals, the wholes are equal.

3. If equals he taken from equals, the remainders are eqvxil,

4. If equals he added to unequals, the wholes are unequal,
the greater sum heing that which includes the greater of the
unequals,

5. If equals he taken from unequals, the remainders are
unequal, the greater remainder heing that which is left from the
greater of the unequals,

6. Things which are double of the same thing, or of equal
things, are equal to one another,

7. Things which are halves of the same thing, or of equal
things, are eqv/il to one another.

9.* The whole is greater than its part.

* To preserve the classification of general and geometrical axioms,
we have placed Euclid^s ninth axiom before the eighth.

Geometrical Axioms.

8. Magnitudes which can he made to coincide with one
another, are equal.

10. Two straight lines cannot enclose a space.

11. All right angles are equal.

12. If a straight line meet two straight lines so as to make
(lie interior angles on one side of it together less than two right
angles, these straight lines mil meet if continually produced on
the side on which are the angles which a/re together less than tvx>
right angles.

GfiOMETRtCAL AXIOMS.

That ifi to say, if the two straight A
lines AB and CD are met by the straight
line EH at F and G, in such a way that
the angles BFG, DGF are together less
than two right angles, it is asserted that
AB and CD will meet if continually
produced in the direction of B and D.

Notes on the Axioms.

1. The necessary characteristics of an Axiom are

(i) That it should be self -evident ; that is, that its truth should
be immediately accepted without proof.

(ii) That it should \}q fundamerUcU ; that is, that its truth should
not be derivable from any other truth more simple than itself.

(iii) That it should supply a basis for the establishment of further
truths.

These characteristics may be summed up in the following defini-
tion.

Definition. An Aziom is a self-evident truth, which neither
requires nor is capable of proof, but which serves as a foundation
for future reasoning.

2. Euclid's Axioms may be classified as general and 'geometrical.

General Axioms apply to magnitudes of all kinds. Geometrical
Axioms refer specially to geometrical magnitudeSy as lines, angles,
and figures.

3. Axiom 8 is Euclid's test of the equality of two geometrical
magnitudes. It implies that any line, angle, or figure, may be
tp.ken up from its position, and without change in size or form,
laid down upon a second line, angle, or figure, for the purpose of
comparison, and it states that two such magnitudes are equal when
one can be exactly placed over the other without overlapping.

This pro(^ess is called superposition, and the first magnitude is
said to be applied to the other.

4. Axiom 12 has been objected to on the double ground that it
cannot be considered self-evident, and that its truth may be deduced
from simpler principles. It is employed for the first time in the 29th
Proposition of Book I. , where a short discussion of the difficulty will
be found. ^

10 EUCLID'S ELEMENTS.

Introductory.

I. Little is known of Euclid beyond the fact that he lived about
three centuries before Christ (325-285) at Alexandria, where he
became famous as a writer and teacher of Mathematics.

Among the works ascribed to him, the best known and most
important is The Ehmenta^ written in Greek, and consisting of
Thirteen Books. Of these it is now usual to read Books I. -IV. and
VI. (which deal with Plane Geometry), together with parts of
Books XI. and XII. (on the Geometry of Solids). The remaining
Books deal with subjects which belong to the theory of Arithmetic,

2. Plane Geometry deals with the properties of all lines and
figures that may be drawn upon a plane surface.

Euclid in his first Six Books confines himself to the properties of
straight lines, rectilineal figures, and circles.

3. The subject is divided into a number of separate discussions^
called propositions.

Propositions are of two kinds, Problems and Theorems.

A Problem proposes to perform some geometrical construction,
such as to draw some particular line, or to construct some required
figure.

A Theorem proposes to prove the truth of some geometrical
statement.

4. A Proposition consists of the following parts :

The General Enunciatiorif the Particular EnuncicUion, the Con-
atruction, and the Proof.

(i) The General Enunciation is a preliminary statement, de-
scribing in general terms the purpose of the proposition.

(ii) The Particular Enunciation repeats in special terms the
statement already made, and refers it to a diagram, which enables
the reader to follow the reasoning more easily.

(iii) The Construction then directs the drawing of such straight
lines and circles as may be required to effect the purpose of a
problem, or to prove the truth of a theorem.

(iv) The Proof shews that the object proposed in a problem has
been aoconjplished, or that the property stated in a theorem is true.

INTRODUCTORY. 11

5. Euclid's reasoning is said to be Deductive, because by a con-
nected chain of argument it dedtices new truths from truths already
proved or admitted. Thus each proposition, though in one sense
complete in itself, is derived from the Postulates, Axioms, or former
propositions, and itself leads up to subsequent propositions.

6. The initial letters Q.E.F., placed at the end of a problem,
stand for Quod erat Faciendum, which was to be daiie.

The letters q.e.d. are appended to a theorem, and stand for Quod
erat Demonstrandum, which was to he proved,

7. A Corollary is a statement the truth of which follows readily
from an established proposition ; it is therefore appended to the
proposition as an inference or deduction, which usually requires no
further proof.

8. The attention of the beginner is drawn to the special use of
the future tense in the Particular Enunciations of Euclid's pro-
positions.

The future is only used in a statement of which the truth is
ahovt to he proved. Thus : ** The triangle ABC shall be equUateral"
means that the triangle has yet to he proved equilateral. While.
" The triangle ABC is equilateral " means that the triangle has
already heen proved (or given) equilateral.

9. The following symbols and abbreviations may be employed in
writing out the propositions of Book I., though their use is not
recommended to beginners.

for

therefore,

par* (or

) for parallel.

^ ,,

is, or are, equal

to.

par"*

„ parallelogram,

^ a

angle.

sq.

,, square,

rt. z „

right angle.

rectil.

„ rectilineal.

A M

triangle.

st. line

„ straight line,

perp. „

perpendicular.

pt.

,, point ;

and all obvious contractions of words, such as opp., adj., diag., etc.,
for opposite, adjacent, diagonal, etc.

12 EUCLID'S ELEMENTa

SECTION I.

Proposition 1. Problem.

To describe an equilateral triangle on a given finite
straight line.

Let AB be the given straight line.
It is required to describe an equilateral triangle on AB.

Construction. With centre A, and radius AB, describe
the circle BCD. Post. 3.

With centre B, and radius BA, describe the circle ACE.

Post. 3.

From the point C at which the circles cut one another,

draw the straight lines CA and CB to the points A and B.

Post, 1.
Then shall the triangle ABC he equilateral.

Proot Because A is the centre of the circle BCD,

therefore AC is equal to AB. Def. 15,

And because B is the centre of the circle ACE,

therefore BC is equal to AB. Def. 15.

Therefore AC and BC are each equal to AB.
But things which are equal to the same thing are equal
to one another. Ax. 1.

Therefore AC is equal to BC.
Therefore AC, AB, BC are equal to one another.
Therefore the triangle ABC is equilateral ;
and it is described on the given straight line AB. Q.E.F,

BOOK I. PROP. 2. 13

Proposition 2. Problem.

From a given point to draw a straight line equal to a given
straight line.

Let A be the given point, and BC the given straight line.
It is required to draw from A a straight line equal to BC.

Construction. Join AB ; Post, 1.

and on AB describe an equilateral triangle DAB. I. 1.
With centre B, and radius BC, describe the circle CGH.

Post. 3.
Produce DB to meet the circle CGH at G. Post. 2.
With centre D, and radius DG, describe the circle GKF.

Produce DA to meet the circle GKF at F. Post. 2.
Then AF shall be equal to BC.

Proot Because B is the centre of the circle CGH,

therefore BC is equal to BG. Def. 15.

And because D is the centre of the circle GKF,

therefore DF is equal to DG. Def. 15.

And DA, a part of DF, is equal to DB, a part of DG ; Def 24.
therefore the remainder AF is equal to the remainder BG.

AX. o.

But BC has been proved equal to BG ;
therefore AF and BC are each equal to BG.
And things which are equal to the same thing are equal
to one another. Ax. 1.

Therefore AF is equal to BC ;
and it has been drawn from the given poiwt ^. c^:^^.

14 BUCLID^S ELEMENTS.

Proposition 3. Problem.

From the greater of two given straight lines to cut off a 'part
eqtLol to the less.

Let AB and C be the two given straight lines, of which
AB is the greater.

It is required to cut off from AB a part equal to C.

Construction. From the point A draw the straight line

AD equal to C ; I. 2.

and with centre A and radius AD, describe the circle DEF,

cutting AB at E. Post 3.

Then AE shall he equal to C.

Proof. Because A is the centre of the circle DEF,

therefore AE is equal t(J AD. JDef 15.

But C is equal to AD. Constr.

Therefore AE and C are each equal to AD.

Therefore AE is equal to C ; Ax, 1.

and it has been cut off from the given straight line AB.

Q.E.P,

EXERCISES ON PROPS. 1-3. 15

EXERCISES ON PROPOSITIONS 1 TO 3.

1. If the two circles in Proposition 1 cut one another again at
F, prove that AFB is an equilateral triangle.

2. If the two circles in Proposition 1 cut one another at C and
F, prove that the figure ACBF is a rhombus.

3. AB is a straight line of given length : shew how to draw from
A a line double the length of AB.

4. Two circles are drawn with the same centre O, and two
radii OA, OB are drawn in the smaller circle. If OA, OB are pro-
duced to cut the outer circle at D and E, prove that AD = BE.

5. AB is a straight line, and P, Q, are two points, one on each
side of AB. Shew how to find points in AB, whose distance from P
is equal to PQ. How many such points will there be ?

6. In the figure of Proposition 2, if AB is equal to BC, shew
that D, the vertex of the equilateral tria?ngle, will fall on the circum-
ference of the circle CGH.

7. In Proposition 2 the point A may be joined to either ex-
tremity of BC. Draw the figure, and prove the proposition in the
case when A is joined to C.

8. On a given straight line AB describe an isosceles triangle
having each of its equal sides equal to a given straight line PQ.

9. On a given base describe an isosceles triangle having each of
its equal sides double of the base.

10. In a given straight line the points A, M, N, B are taken in
order. On AB describe a triangle AoC, such that the side AC may
be equal to AN, and the side BU to BM.

NOTE ON PROPOSITIONS 2 AND 3.

Propositions 2 and 3 are rendered necessary by the restriction
tacitly imposed by Euclid, that compasses shall not be used to
transfer distances. [See Notes on the Postulates.]

In carrying out the construction of Prop. 2 the point A may be
joined to either extremity of the line BC ; the equilateral triangle
may be described on either side of the line so drawn ; and the sides
of the equilateral triangle may be produced in either direction. Thus
there are in general 2x2x2, or eight , possible constructions. The
student should exercise himself in drawing the various fi^uxe^ \)c\a.\»
may arise.

16 euclid's elements.

Proposition 4. Theorem.

If two triangles have two sides of the one equal to two sides
of the other, each to each, and have also the angles contained
by those sides equal, then the triangles shall be equal in all
respects ; that is to say, their bases or third sides shall be equal,
and their remaining angles shall be equal, each to each, namely
those to which the equal sides are opposite; and the triangles
shall be equal in area.

Let ABC, DEF be two triangles, in which
the side AB is equal to the side DE,
the side AC is equal to the side DF, and
the contained angle BAC is equal to the contained angle EDF,

Then (i) the base BC shall be equal to the base EF ;

(ii) the angle ABC shall be equal to the angle DEF ;
(iii) the angle ACB shall be equal to the angle DFE ;
(iv) the triangle ABC shall be equoCl to the tiiangle
DEF in area.

Proof. If the triangle ABC be applied to the triangle DEF,

so that the point A may lie on the point D,
and the straight line AB along the straight line DE;

then because AB is equal to DE, Hyp.

therefore the point B must coincide with the point E.
And because AB falls along DE,
and the angle BAC is equal to the angle EDF, Hyp,
therefore AC must fall along DF.
And because AC is equal to DF, Hyp.

therefore the point C must coincide with the point F.

Then since B coincides with E, and C with F,
therefore the base BC must coincide with the base

BOOK I. PROP. 4. 17

for if not, two straight lines would enclose a space ; which

is impossible. Ax. 10.

Thus the base BC coincides with the base EF, and is

therefore equal to it. Ax, 8.

And the remaining angles of the triangle ABC coincide

with the remaining angles of the triangle DEF, and are

therefore equal to them ;

namely, the angle ABC is equal to the angle DEF,
and the angle ACB is equal to the angle DFE.
And the triangle ABC coincides with the triangle DEF,
and is therefore equal to it in area. Ax, 8.

That is, the triangles are equal in all respects. q.e.d.

Note. The sides and angles of a triangle are known as its six
parts, A triangle may also be considered in regard to its area.

Two triangles are said to be equal in aU respects, or identically
equal, when the sides and angles of one are respectively equal to
the sides and angles of the other. We have seen that such triangles
may be made to coincide with one another by superposition^ so that
they are also equal in area. [See Note on Axiom 8.]

[It will be shewn later that triangles can be equal in area without
being equal in their several parts ; that is to say, triangles can have
the same area without having the same sJiape.]

EXERCISES ON PROPOSITION 4.

1. ABCD is a square : prove that the diagonals AC, BD are equal
to one another.

2. ABCD is a square, and L, M, and N are the middle points of
AB, BC, and CD : prove that

(i) LM = MN. (ii) AM = DM.

(iii) AN=AM. (iv) BN = DM.

[Draw a separate figure in each case.]

3. ABC is an isosceles triangle : from the equal sides AB, AC
two equal parts AX, AY are cut off, and BY and CX are joined.
Prove that BY = ex.

4. ABCD is a quadrilateral having the opposite sides BC, AD
equal, and also the angle BCD equal to the angle ADC : i^tON^ tW\i
BD is equal to AC.

18 Euclid's elements.

Proposition 5. Theorem.

The angles at the base of an isosceles tnangle are equal to
one another ; and if the equal sides be jyfoduced^ the angles oil
the other side of the base shall also be equal to one another.

Let ABC be an isosceles triangle, in which
the side AB is equal to the side AC,
and let the straight lines AB, AC be produced to D and E.

Then (i) the angle ABC shall be equal to th^ angle ACB;
(ii) the angle CBD sJuill be equal to the angle BCE.

Construction. In BD take any point F ;

and from AE cut off a part AG equal to AF. I. 3.

Join FC, GB.

Proof. Then in the triangles FAC, GAB,

(FA is equal to GA, Constr.

and AC is equal to AB, Hyp.

also the contained angle at A is common to the
two triangles :
therefore the triangle FAC is equal to the triangle GAB in
all respects; I. 4.

that is, the base FC is equal to the base GB,
and the angle ACF is equal to the angle ABG,
also the angle AFC is equal to the angle AGB.

Again, because AF is equal to AG,
and AB, a part of AF, is equal to AC, a part of AG ; Hyp.
therefore the remainder BF is equal to the remainder CQ.

BOOK I. PROP. 5. 19

Then in the two triangles BFC, CGB,

BF is equal to CG, Proved,

T> i and PC is equal to GB, Proved,

also the contained angle BFC is equal to the

contained angle CGB, Proved.

therefore the triangle BFC is equal to the triangle CGB in

all respects ; i. 4.

so that the angle FBC is equal to the angle GCB,

and the angle BCF to the angle CBG.

Now it has been shewn that the angle ABG is equal to the
angle ACF,

and that the angle CBG, a part of ABG, is equal to the angle
BCF, a part of ACF ;

therefore the remaining angle ABC is equal to the remain-
ing angle ACB ; Ax, 3.
and these are the angles at the base of the triangle ABC.

Also it has been shewn that the angle FBC is equal to the

angle GCB ;
and these are the angles on the other side of the base. Q.E.D.

Corollary. Hence if a triangle is equilateral it is also
equiangular.

Note. The difficulty which be-
ginners find with this proposition
arises from the fact that the triangles
to be compared overlap one another
in the diagram. This difficulty may
be diminished by detaching eacjfi
pair of triangles from the rest of the
figure, as sheMrn in the margin.

20 EUCLID'S ELEMENTS.

Proposition 6. Theorem.

If two angles of a triangle he equal to one another^ then the
sides also which subtend, or are opposite to, the equal angles,
shall be equal to one another.

Let ABC be a triangle, in which
the angle ABC is equal to the angle ACB.

Then shall the side AC be equal to tJie side AB.

Construction. For if AC be not equal to AB,
one of them must be greater than the other.
If possible, let AB be the greater ;
and from it cut off BD equal to AC. I. 3.

Join DC.

Proot Then in the triangles DBC, ACB,

(DB is equal to AC, Constr,

and BC is common to both,
also the contained angle DBC is equal to the
contained angle ACB ; Hi/p,

therefore the triangle DBC is equal to the triangle ACB
in area, I. 4.

the part equal to the whole ; which is absurd. Az, 9.

Therefore AB is not unequal to AC ;

that is, AB is equal to AC. Q.E.D.

Corollary. Hence if a triangle is equiangular it is also
equilateral.

ON BOOK I. PROPS. 6 AND 6. 21

NOTE ON PROPOSITIONS 5 AND 6.

The enunciation of a theorem consists of two clauses. The first
clause tells us what we are to assume, and is called the hypothesis ;
the second tells us what it is required to prove, and is called the
conclusion.

For example, the enunciation of Proposition 5 assumes that in a
certain triangle ABC the aide AB=^Ae side AC : this is the hypothesis.
From this it is required to prove that the angle kSO = the angle ACB :
this is the conclusion.

If we interchange the hypothesis and conclusion of a theorem,
we enunciate a new theorem which is called the converse of the
first.

For example, in Prop. 5

it is assumed that AB = AC ;

it is required to jyrove that the angle ABC = the angle ACB.

Now in Prop. 6

it is assumed that the angle ABC = the angle ACB ;\
it is required to prove that AB=AC. j

Thus we see that Prop. 6 is the converse of Prop. 5 ; for the hypo-
thesis of ea^h is the conclusion of the other.

In Proposition 6 Euclid employs for the first time an indirect
method of proof frequently used in geometry. It consists in shewing
that the theorem cannot he untrue ; since, if it were, we should be
led to some impossible conclusion. This form of proof is known as
Reductio ad- Absurdum, and is most commonly used in demonstrat-
ing the converse of some foregoing theorem.

The converse of all true theorems are not themselves necessarily
true. [See Note on Prop 8.]

EXERCISES ON PROPOSITION 5.

1. ABCD is a rhombus, in which the diagonal BD is drawn :
shew that ^jj ^j^^ ^^^^ j^^^^ ^j^e ^j^gj^ j^^^ .

(ii) the angle CBD = the angle CDB ;
(iii) the angle ABC = the angle ADC.

2. ABC, DBC are two isosceles triangles drawn on the same
base BC, but on opposite sides of it : prove (by means of i. 5) that
the angle ABD=the angle ACD.

3. ABC, DBC are two isosceles triangles drawn on the same
base BC and on the same side of it : employ i. 5 to prove that the
angle ABD=the angle ACD.

22 EtJCLID*S ELEMENTS

Proposition 7. Theorem.

On the same base^ and on the same side of it, there cannot be
two triangles having their sides which are terminated at one
extremity of the base equal to one another, and likewise those
which are terminated at the other extremity equal to one another.

If it be possible, on the same base AB, and on the same
side of it, let there be two triangles ACB, ADB in which

the side AC is equal to the side AD,
and also the side BC is equal to the side BD.

Case I. When the vertex of each triangle is without
the other triangle.

ConstructioD. Join CD.

Proof. Then in the triangle ACD,

because AC is equal to AD, Hyp.

therefore the angle ACD is equal to the angle ADC. I. 5.

But the whole angle ACD is greater than its part, the

angle BCD ;
therefore also the angle ADC is greater than the angle BCD;
still more then is the angle BDC greater than the angle
BCD.

Again, in the triangle BCD,
because BC is equal to BD, Hyp,

therefore the angle BDC is equal to the angle BCD : i. 5.
but it was shewn to be greater ; which is impossible.

BOOK I. PROP. 7. 23

Case II. When one of the vertices, as D, is within
the other triangle ACB.

Construction. As before, join CD ;

and produce AC, AD to E and F.

Proofc Then in the triangle ACD,

because AC is equal to AD, Hyp,

therefore the angle ECD is equal to the angle FDC,
these being the angles on the other side of the base. i. 5.

But the angle ECD is greater than its part, the angle BCD ;
therefore the angle FDC is also greater than the angle

BCD:
still more then is the angle BDC greater than the angle
BCD.

Again, in the triangle BCD,
because BC is equal to BD, Hyp,

therefore the angle BDC is equal to the angle BCD : i. 5.
but it has been shewn to be greater ; which is impossible.

The case in which the vertex of one triangle is on a
side of the other needs no demonstration.

Therefore AC cannot be equal to AD, and at the same
timey BC equal to BD. q.e.d.

Note. The sides AC, AD are called conterminous sides ; similarly
the sides BC, BD arc conterminous.

24 EUCLID'S ELEMENTS.

Proposition 8. Theorem.

.»

If two triangles hive two sides of the one equal to two sides of
the other, each to each, and have likewise their bases equal, then
the angle which is contained by the two sides of the one shall be
equal to tlie angle which is contained by the two sides of the
other

D G

Let ABC, DEF be two triangles, in which

the side AB is equal to the side DE,

the side AC is equal to the side DF,

and the base BC is equal to the base EF.

Then shall the angle BAC be equal to the angle EDF.

Proot If the triangle ABC be applied to the triangle DEF,
so that the point B falls on the point E,
and the base BC along the base EF ;
then because BC is equal to EF, Hyp.

therefore the point C must coincide with the point F.

Then since BC coincides with EF,
it follows that BA and AC must coincide with ED and DF :
for if they did not, but took some other position, as EG, GF,
then on the same base EF, and on the same side of it, there
would be two triangles EDF, EGF, having their conterminous
sides equal : namely ED equal to EG, and FD equal to FG.

But this is impossible. i. 7.

Therefore the sides BA, AC coincide with the sides ED, DF.
That is, the angle BAC coincides with the angle EDF, and is
therefore equal to it. Ax. 8.

Q.E.D.

NOTE AND EXERCISES ON BOOK I. PROP 8. 25

Note 1. In this Proposition the three sides of one triangle are
given equal respectively to the three sides of the other ; and from
this it is shewn that the two triangles may be made to coincide with
one another.

Hence we are led to the following important Corollary.

Corollary. If in two triangles the three sides of the one
are equal to the three sides of the other, each to each, then the
triangles are equal in all respects,

[An alternative proof, which is independent of Prop. 7, will be
found on page 26.]

Note 2. Proposition 8 furnishes an instance of a true theorem
of which the converse is not necessarily true.

It is proved above that */ the sides of one triangle are severatlj;
equal to the sides of another, then the angles of the first triangle
are severally equal to the angles of the
second.

The converse of this enunciation would
be as follows : If the angles of one tri-
angle are severally equal to the angles of
another y then the sides of the first triangle
are equal to the sides of the second.

But this, as the diagram in the margin shews, is by no means
necessarily true.

EXERCISES ON PROPOSITION 8.

1. Shew (by drawing a diagonal) that the opposite angles of a
rhombus are equal.

2. If ABCD is a Quadrilateral, in which AB = CD and AD = CB,
prove that the angle ADC = the angle ABC.

3. If ABC and DBC are two isosceles triangles drawn on the
same base BC, prove (by means of i. 8) that the angle ABD = the
angle ACD, taking (i) the case where the triangles are on the same
side of BC, (ii) the case where they are on opposite sides of BC.

4. If ABC, DBC are two isosceles triangles drawn on opposite
sides of the same base BC, and if AD be joined, prove that each of
the angles BAC, BDC will be divided into two equal parts.

5. If in the figure of Ex. 4 the line AD meets BC in E, prove
that BE = EC.

EUCLID S ELEMENTS.

Proposition 8. Alternative Proof.

Let ABC and DEF be two triangles, which have the sides BA, AC
equal respectively to the sides ED, DF, and the base BC equal to
the base EF.

Then shall the amjle BAC he equal to the angle EDF.

For apply the triangle ABC to the triangle DEF, so that B may
fall on E, and BC along EF, and so that the point A may be on the
side of EF remote from D ;

then C must fall on F, since BC is equal to EF.
Let GEF be the new position of the triangle ABC.

Join DG.

Case L When DG intersects EF.

Then because ED = EG,

the angle EDG=the angle EGD.

Again because FD = FG,

the angle FDG = the angle FGD.

Hence tlie whole angle EDF = the whole angle EGF ; Ax, 2,
that is, the angle EDF — the angle BAG.

Two cases remain which may be dealt with in a similar manner :
namely,

I. 5.

Case II. When DG meets EF
produced.

B C

Case III. When one pair of
sides, as DF, FG are in one straight q
line.

QUESTIONS FOR REVISION. 27

QUESTIONS AND EXERCISES FOR REVISION.

1. Define CLdjacerU angleSf a right angle, vertically opposite angles,

2. Explain the words enunciation^ hypothesis, conclusion.

3. Distinguish between the meanings of the following statements :

(i) then AB is equal to PQ ;
(ii) then AB shoM he equal to PQ.

4. When are two theorems said to be coriiverse to one another.
Give an example.

5. Shew by an example that the converse of a true theorem is
not itself necessarily true.

6. What is a corollary ? Quote the corollary to Proposition 5 ;
and shew how its truth follows from that proposition.

7. Name the six parts cf a triangle. When are triangles said
to be eqxiaX in all respects ?

8. What do you understand by the expression geometrical
magnitudes ? Give examples ?

9. What is meant by superposition ? Explain the test by which
Euclid determines if two geometrical magnitudes are equal to one
another. Illustrate by an example.

10. Quote and explain the third postulate. What restrictions
does Euclid impose on the use of compasses, and what problems are
thereby made necessary ?

11. Define an axiom. Quote the axioms referred to (i) in Pro-
position 2 ; (ii) in Proposition 7.

12. Prove by the method of superposition that two squares are
equal in area, if a side of one is equal to a side of the other.

13. Two quadrilaterals ABCD, EFGH have the sides AB, BO,
CD, DA equal respectively to the sides EF, FG, GH, HE, and have
also the angle BAD equal to the angle FEH. Shew that the figures
may be made to coincide with one another.

14. AB, AC are the equal sides of an isosceles triangle ABC ;
and L, M, N are the middle points of AB, BC, and CA respectively :
prove that

(i) LM = MN. (ii) BN=CL.

(iii) the angle ALM = the angle AN M ,

28 EUCLID'S ELEMENTS.

Proposition 9. Problem.

To bisect a given rectilineal angle, that is, to divide it into
two equal parts.

Let BAG be the given angle.
It is required to bisect the angle BAG.

Construction. In AB take any point D ;

and from AG cut off AE equal to AD. I. 3.

Join DE ;
and on DE, on the side remote from A, describe an equi-
lateral triangle DEF. I. 1.

Join AF.
Then shall the straight line AF bisect the angle BAG.

Proot For in the two triangles DAF, EAF,

{DA is equal to EA, Constr.

and AF is common to both ;
and the third side DF is equal to the third side
EF ; Def, 24.

therefore the angle DAF is equal to the angle EAF. I. 8.

Therefore the given angle BAG is bisected by the straight
line AF. Q.KF.

EXERCISES.

1. If in the above figure the equilateral triangle DFE were de-
scribed on the same side of DE as A, what different cases would
arise ? And under what circumstances would the construction fail ?

2. In the same figure, shew that AF also bisects the angle DFE.

3. Divide an angle into four equal parts.

BOOK I. PROP. 10. 29

Proposition 10. Problem.

To Used a given finite straight line, that is, to divide it into
two equal parts.

Let AB be the given straight line.
It is required to divide AB into two equal parts,

Constr. On AB describe an equilateral triangle ABC; I. 1.
and bisect the angle ACB by the straight line CD, meeting
AB at D. I. 9.

Then shall AB he bisected at the point D.

Proof. For in the triangles ACD, BCD,

{AC is equal to BC, Def, 24.

and CD is common to both :
also the contained angle ACD is equal to the con-
tained angle BCD ; Constr,
therefore the triangle ACD is equal to the triangle BCD in
all respects : i. 4.
so that the base AD is equal to the base BD.

Therefore the straight line AB is bisected at the point D.

Q.E.F.

EXERCISES.

1. Shew that the straight line which bisects the vertical angle
of an isosceles triangle, also bisects the base.

2. On a given base describe an isosceles triangle such that the
sum of its equal sides may be equal to a given straight linev

30 euclid's elements.

Proposition 11. Problem.

To draw a straight line at right angles to a given straight
line, from a given point in the same.

Let AB be the given straight line, and O the given
point in it.

It is required to draw from C a straight line at right angles
to AB.

Construction. In AC take any point D,

and from CB cut off CE equal to CD. I. 3.

On DE describe the equilateral triangle DFE. L 1.

Join CF.
Then shall CF he at right angles to AB.

Proof. For in the triangles DCF, ECF,

(DC is equal to EC, Constr.

and CF is common to both ;
and the third side DF is equal to the third side
EF : Def 24.

therefore the angle DCF is equal to the angle ECF : L 8.
and these are adjacent angles.

But when one straight line, standing on another, makes

the adjacent angles equal, each of these angles is called a

right angle ; Def 10.

therefore each of the angles DCF, ECF is a right angle.

Therefore CF is at right angles to AB,

and has been drawn from a point C in it. Q.E.F,

EXERCISE.

In the figure of the above proposition, shew that any point in
PC, or FC produced, is equidistant from D and E.

book i. prop. 12. 31

Proposition 12. Problem.

To draw a straight line perpmdicvlar to a given straight line
of unlimited lengthy from a given point without it.

Let AB be the given straight line of unlimited length,
and let C be the given point without it.

It is required to draw from C a straight line pmpendicular
to AB.

CJonstruction. On the side of AB remote from C take
any point D ;
and with centre C, and radius CD, describe the circle FDG,
cutting AB at F and G

Bisect FGat H; 1. 10.

and join CH.
Then shall OH he perpendicular to AB.
Join CF and CG.

Proot Then in the triangles FHC, GHC,

(FH is equal to GH, Constr.

and HC is common to both ;
and the third side CF is equal to the third side
CG, being radii of the circle FDG ; Def 15.

therefore the angle CHF is equal to the angle CHG ; I. 8.
and these are adjacent angles.

But when one straight line, standing on another, makes

the adjacent angles equal, each of these angles is called a

right angle, and the straight line which stands on the

other is called a perpendicular to it. Def. 10.

Therefore CH is perpendicular to AB,

and has been drawn from the point C without it. q.e.f.

Note. The line AB must be of unlimited length, that is, capable
of production to an indefinite length in eitlier direction, to ensure
its being intersected in two points by the circle F DG.

32 EUCLID'S ELEMENTS.

QUESTIONS AND EXERCISES FOR REVISION.

1. Distinguish between a problem and a theorem,

2. When are two figures said to be identiccUly equal ? Under
what conditions has it so far been proved that two triangles are
identically equal ?

3. Explain the method of proof known as Reductio ad A hsurdum.
Quote the enunciations of the propositions in which this method has
so far been used.

4. Quote the corollaries of Propositions 5 and 6, and shew that
each is the converse of the other.

5. What is meant by saying that Euclid's reasoning is deductive ?
Shew, for instance, that the proof of Proposition 5 is a deductive
argument.

6. Two forts defend the mouth of a river, cmfi on each side ;
the forts are 4000 yards apart, and their guns have a range of 3000
yards. Taking one inch to represent a length of 1000 yards, draw a
diagram shewing what part of the river is exposed to the fire of both
forts.

7. Define the perimeter of a rectilineal figure. A square and an
equilateral triangle each have a perimeter of 3 feet : compare the
lengths of their sides.

8. Shew how to draw a rhombus each of whose sides is equal to
a given straight line PQ, which is also to be one diagonal of the
figure.

9. A and B are two given points. Shew how to draw a rhomuus
having A and B as opposite vertices, and having each side equal to a
given line PQ. Is this always possible ?

10. Two circles are described with the same centre O ; and two
radii OA, OB are drawn to the inner circle, and produced to cut the
outer circle at D and E : prove 'that

(i) DB = EA;

(ii) the angle BAD = the angle ABE;
(iii) the angle ODB = the angle OE A,

EXERCISES ON BOOK I. PROPS. 1-12. 33

EXERCISES ON PROPOSITIONS 1 TO 12.

1. Shew that the straight line which joins the vertex of an
isosceles triangle to the middle point of the base is perpendicular
to the base.

2. Shew that the straight lines which join the extremities of
the l»se of an isosceles triangle to the middle points of the opposite
sides, are equal to one another.

3. Two given points in the base of an isosceles trianele are equi-
distant from the extremities of the base : shew that they are also
equidistant from the vertex.

4. If the opposite sides of a quadrilateral are equal, shew that
the opposite angles are also equal.

5. Any two isosceles triangles XAB, YAB stand on the same
base AB : shew that the angle a AY is equal to the angle XBY ; and
if XY be joined, that the angle AXY is equal to the angle BXY.

6. Shew that the opposite angles of a rhombus are bisected by
the diagonal which joins them.

7. Shew that the straight lines which bisect the base angles of
an isosceles triangle form with the base a triangle which is also
isosceles.

8. ABC is an isosceles triangle having AB equal to AC ; and the
Angles at B and C are bisected by straiight lines which meet at O :
shew that OA bisects the angle BAC.

9. Shew that the triangle formed by joining the middle points
of the sides of an equilateral triangle is also equilateral.

10. The equal sides BA, CA of an isosceles triangle BAC are pro-
duced beyond the vertex A to the points E and F, so that AE is
equal to AF ; and FB, EC are joined : shew that FB is equal to EC.

11. Shew that the diagonals of a rhombus bisect one another at
right angles.

12. In the equal sides AB, AC of an isosceles triangle ABC two

g>ints X and Y are taken, so that AX is equal to AY ; and CX and
Y are drawn intersecting in O : shew that

(i) the triangle BOC is isosceles ;
(ii) AG bisects the vertical angle BAC ;
(iii) AG, if produced, bisects BC at right angles.

13. Describe an isosceles triangle, having given the base and the
length of the perpendicular drawn from the vertex to the base.

14. In a given straight line find a point that is equidistant from
two given points. In what case is this impossible^.

H.&E. o

EUCLID'S ELEMENTS.

Proposition 13. Theorem.

The adjacent angles which one straight line makes with
another straight line, on one side of it, are either two right angles^
or are together equal to two right angles.

Let the straight line AB meet the straight line DC.
Then the adjacent angles DBA, ABC shall he either two right
angles, or together equal to two right angles.

Case I. For if the angle DBA is equal to the angle
ABC, each of them is a right angle. Def, 10.

Case II. But if the angle DBA is not equal to the
angle ABC,

from B draw BE at right angles to CD. I. 11.

Proof. Now the angle DBA is made up of the two
angles DBE, EBA;

to each of these equals add the angle ABC ;
then the two angles DBA, ABC are together equal to the
three angles DBE, EBA, ABC. Ax, 2.

Again, the angle EBC is made up of the two angles EBA,
ABC;

to each of these equals add the angle DBE ;
then the two angles DBE, EBC are together equal to the
three angles DBE, EBA, ABC. Ax, 2.

But the two angles DBA, ABC have been shewn to be equal

to the same three angles ;
therefore the angles DBA, ABC are together equal to the

angles DBE, EBC. Ax, 1.

But the angles DBE, EBC are two right angles ; Constr,

therefore the angles DBA, ABC are together equal to two

right angles. Q.E.D.

BOOK I. PROP. 13. 35

DEFINITIONS.

(1) The complement of an acute angle is its defect from a right
angle, that is, the angle by which it falls short of a right angle.

Thus two angles are complementary, when their sum is a right
angle.

(ii) The supplement of an angle is its defect from two right
angles, that is, the angle by which it falls short of two right angles.

Thus two angles are supplementary, when their sum is two
right angles.

Ck)BOLLABY. Angles which are complementary or supplementary
to the same angle ore equal to one another.

EXERCISES.

1. If the two exterior angles formed by producing a side of a
triangle both ways are equal, shew that the triangle is isosceles.

2. The bisectors of the adjacent angles which one straight line
makes with another contain a right angle.

Note In the adjoining diagram AOB is a given angle ; and one
of its arms AO is produced to C : the
adjacent angles AOd, BOC are bisected
by OX, OY.

Then OX and OY are called respect-
ively the internal and external bisectors
of the angle AOB.

Hence Exercise 2 may be thus
enunciated :

The internal and external "bisectors of an angle are at right angles
to one another,

3. Shew that the angles AOX and COY are complementary.

4. Shew that the angles BOX and COX are supplementary ; and
also that the angles AOY and BOY are Bupplemeiit8.ry.

36 euclid's elements.

Proposition 14. Theorem.

Ify at a poitU in a straight line, tvx> other straight lines, cm
opposite sides of it, make the adjacerU angles together equal to
two right angles, then these two straight lines shall he in one
and the same straight line,

At the point B in the straight line AB, let the two
straight lines BC, BD, on the opposite sides of AB, make
the adjacent angles ABC, ABD together equal to two right
angles.

Then BD shall he in the same straight line with BC.

Proof. For if BD be not in the same straight line with BC,
if possible, let BE be in the same straight line with BC.

Then because AB meets the straight line CBE,
therefore the adjacent angles CBA, ABE are together equal

to two right angles. L 13.

But the angles CBA, ABD are also together equal to two

right angles. Eyp.

Therefore the angles CBA, ABE are together equal to the

angles CBA, ABD. Ax, 11.

From each of these equals take the common angle CBA ;
then the remaining angle ABE is equal to the remaining angle

ABD 'y the part equal to the whole ; which is impossible.

Therefore BE is not in the same straight line with BC.
And in the same way it may be shewn that no other
line but BD can be in the same straight line with BC.

Therefore BD is in the same straight line with BC. Q.E.D.

exercise.

ABCD is a rhombus ; and the diagonal AC is bisected at O. If O
is joined to the angular points B and D ; shew that OB and CD are
in one straight line.

BOOK I. PROP. 15. 37

Proposition 15. Theorem.

If two straight lines intersect one another, then the vertically
opposite amgles shall he equal.

Let the two straight lines AB, CD cut one another at
the point E.

Then (i) the angle AEC shall he eqml to the angle DEB;
(ii) the angle CEB shall he equal to the angle AED.

Proot Because AE meets the straight line CD,
therefore the adjacent angles CEA, AED are together equal
to two right angles. i. 13.

Again, because DE meets the straight line AB,
therefore the adjacent angles AED, DEB are together equal

to two right angles. i. 13.

Therefore ^e angles CEA, AED are together equal to the

angles AED, DEB.

From each of these equals take the common angle AED ;
then the remaining angle CEA is equal to the remaining
angle DEB. Ax. 3.

In the same way it may be proved that the angle CEB
is equal to the angle AED. Q.E.D.

C!oROLLARY 1. From this it follows that, if two straight
lines cut one a/nother, the four angles so formed are together
equal to fcmr right ambles.

Corollary 2. Consequently, when any number of straight
lines meet at a point, the sum of the angles made hy consecutive
lines is equal to four right angles.

38 Euclid's elements.

Proposition 16. Theorem.

If one side of a triangle he jproducedi then the exterior angle
shall he greater than either of the interior opposite angles.

Because

Let ABC be a triangle, and let BC be produced to D.
Then shall the exterior angle ACD be greater than either of
the interior opposite angles ABC, BAG.

Construction. Bisect AC at E ; I. 10.

Join BE ; and produce it to F, making EF equal to BE. L 3.

Join FC.

Proof. Then in the triangles AEB, CEF,

AE is equal to CE, Constr.

and EB is equal to EF ; Constr,

also the angle AEB is equal to the vertically

, opposite angle CEF; L 15.

therefore the triangle AEB is equal to the triangle CEF in

all respects : L 4.

so that the angle BAE is equal to the angle ECF.

But the angle ECD is greater than its part, the angle ECF ;

therefore the angle ECD is greater than the angle BAE ;

that is, the angle ACD is greater than the angle BAC.

In the same way, if BC be bisected, and the side AC
produced to G, it may be proved that the angle BCQ is
greater than the angle ABC. •

But the angle BCG is equal to the angle ACD : I. 15,
therefore also the angle ACD is greater than the angle ABC.

Q.E.D.

book i. prop. 17. 39

Proposition 17. Theorem.

Any tvm angles of a triangle a/re together less than two
right angles,

Let ABC be a triangle.
Then shall any two of the angles of the triangle ABC he
together less than two right angles.

Construction. Produce the side BC to D.

Proof: Then because BC, a side of the triangle ABC, is

produced to D ;
therefore the exterior angle ACD is greater than the interior

opposite angle ABC. i. 16.

To each of these add the angle ACB :
then the angles ACD, ACB are together greater than the

angles ABC, ACB. Ax, 4.

But the adjacent angles ACD, ACB are together equal to

two right angles. i. 13.

Therefore the angles ABC, ACB are together less than two

right angles.
Similarly it may be shewn that the angles BAC, ACB, as
also the angles CAB, ABC, are together less than two right
angles. q.e.d.

Note. It follows from this Proposition that every triangle must
have at least two actUe angles : for if one angle is obtuse, or a right
angle, each of the other angles must be less than a right angle.

exercises.

1. Enunciate this Proposition so as to shew that it is the con-
verse of Axiom 12.

2. If any side of a triangle is produced both ways, the exterior
angles so formed are together greater than two right angles.

3. Shew how a proof of Proposition 17 may be obtained by
joining each vertex in turn to any point in the opposite side.

40 EUCLID'S ELEMENTS.

Proposition 18. Theorem.

If one side of a triangle be greater than another, then the
angle opposite to the greater side shaU be greater than the angle
opposite to the less.

Let ABC be a triangle, in which the side AC is greater
than the side AB.

Then shall the angle ABC be greater than the angle ACB.

Construction, From AC cut oif a part AD equal to AB. L 3.

Join BD.

Proof. Then in the triangle ABD,

because AB is equal to AD,
therefore the angle ABD is equal to the angle ADB. I. 5.

But the exterior angle ADB of the triangle DCB is
greater than the interior opposite angle DCB, that is,
greater than the angle ACB. I, 16.

Therefore also the angle ABD is greater than the angle ACB;

still more then is the angle ABC greater than the angle

ACB. Q.E.D.

Euclid enunciated Proposition 18 as follows :

The greater side of every triangle has the greater angle opposite
to it,

[This form of enunciation is found to be a common source of diffi-
culty with beginners, who fail to distinguish what is assumed in it
and what is to be proved. If Euclid's enunciations of Props. 18 and
19 are adopted, it is important to remember that in each case the
part of the triangle ^r«^ narned points out the hypothesis.]

book i. prop. 19. 41

Proposition 19. Theorem.

If one angle of a triangle he greater than another, then the
side opposite to the greater angle shall be greater than the side
opposite to the less.

Let ABC be a triangle in which the angle ABC is greater
than the angle ACB.

Then shall the side AC he greater than the side AB.

Proot For if AC be not greater than AB,
it must be either equal to, or less than AB.

But AC is not equal to AB,
for then the angle ABC would be equal to the angle ACB; I. 5.

but it is not. Hyp,

Neither is AC less than AB ;
for then the angle ABC would be less than the angle ACB; 1. 18.

but it is not. Hyp,

That is, AC is neither equal to, nor less than AB.

Therefore AC is greater than AB. Q.E.D.

KoTE. The mode of demonstration used in this Proposition is
known as the Proof by Exhaustion. It is applicable to cases in which
one of certain suppositions must necessarily be true ; and it consists
in shewing that each of these suppositions is false toitJi one exception :
hence the truth of the remaining supposition is inferred.

Euclid enunciated Proposition 19 as follows :

The greater angle of every triangle is subtended by the greater
side, or, has the greater side opposite to it.

[For Exercises on Props. 18 and 19 see page 44."^

42 EUCLID'S ELEMENTS.

Proposition 20. Theorem.

Any two sides of a triangle are together greater than the
third side.

Let ABC be a triangle.
Then shall any two of its sides be together greater than the
third side :

namely, BA, AC, shall be greater than CB ;

AC, CB shall be greater than BA ;
and CB, BA shall be greater than AC.

Construction. Produce BA to D, making AD equal to AC. I. 3.

Join DC.

Prool Then in the triangle ADC,

because AD is equal to AC, Constr.

therefore the angle ACD is equal to the angle ADC. I. 5.

But the angle BCD is greater than its part the angle ACD ;

therefore also the angle BCD is greater than the angle ADC,

that is, than the angle BDC.

And in the triangle BCD,
because the angle BCD is greater than the angle BDC,
therefore the side BD is greater than the side CB. i. 19,

But BA and AC are together equal to BD ;
therefore BA and AC are together greater than CB.

Similarly it may be shewn

that AC, CB are together greater than BA ;

and CB, BA are together greater than AC. Q.E.D.

[For Exercises see page 44.]

book l prop. 21, 43

Proposition 21. Theorem.

If from the ends of a side of a triangle^ there he drawn two
straight lines to a point within the triangle, then these straight
lines shall be less than the other two sides of the triangle, hut
shall contain a greater angle.

Let ABC be a triangle, and from B, C, the ends of the
side BC, let the straight lines BD, CD be drawn to a point
D within the triangle

Then (i) BD and DC shall he together less than BA and AC ;
(ii) the angle BDC shall he greater than the angle BAC.

Construction. Produce BD to meet AC in E.

Prool (i) In the triangle BAE, the two sides BA, AE are
together greater than the third side BE ; i. 20.

to each of these add EC ;
then BA, AC are together greater than BE, EC. Ax, 4.

Again, in the triangle DEC, the two sides DE, EC are to-
gether greater than DC ; I. 20.

to each of these add BD ;
then BE, EC are together greater than BD, DC.

But it has been shewn that BA, AC are together greater
than BE, EC :

still more then are BA, AC greater than BD, DC.

(ii) Again, the exterior angle BDC of the triangle DEC is
greater than the interior opposite angle DEC ; i. 16.

and the exterior angle DEC of the triangle BAE is greater
than the interior opposite angle BAE, that is, than the
angle BAC; I. 16.

still more then is the angle BDC greater than the angle BAC.

44 EUCLID'S ELEMENTS.

EXERCISES.
ON PbOPOSITIONS 18 AUD 19.

1. The hypotenuse is the greatest side of a right-angled triangle.

2. If two angles of a triangle are equal to one another, the
sides also, which subtend the equal angles, are equal to one another.
Prove this [t.c. Prop. 6] indirectly by using the result of Prop. 18.

3. BC, the base of an isosceles triangle ABC, is produced to any
point D ; shew that AD is greater than either of the equal sides.

4. If in a quadrilateral the greatest and least sides are opposite
to one another, then each of the angles adjacent to the least side is
greater than its opposite angle.

5. In a- triangle ABC, if AC is not greater than AB, shew that
any straight line drawn through the vertex A and terminated by the
base BC, is less than AB.

6. ABC is a triangle, in which OB, OC bisect the angles ABC,
ACB respectively : shew that, if AB is greater than AC, then OB is
greater than Ou.

ON Proposition 20.

7. The difference of any two sides of a triangle is less than
the third side.

8. In a quadrilateral, if two opposite sides which are not parallel
are produced to meet one another ; shew that the perimeter of the
greater of the two triangles so formed is greater than the perimeter
of the quadrilateral.

9. The sum of the distances of any point from the three angular
points of a triangle is greater than half its perimeter.

10. The perimeter of a quadrilateral is greater than the sum of
its diagonals.

11. Obtain a proof of Proposition 20 by bisecting an angle by a
straight line which meets the opposite side.

ON Proposition 21.

12. In Proposition 21 shew that the angle BDC is greater than
the angle BAC by joining AD, and producing it towards the base.

13. The sum of the distances of any point within a triangle from
its angular points is less than the perimeter of the triangle.

QUESTIONS ON BOOK I. PROPS. 1-21. 45

QUESTIONS FOR REVISION.

1. Define the complemmt of an angle. When are two angles said
to be s^'p^lefm&nlary'i Shew that two angles which are supple-
mentary to the same angle are equal to one another.

2. What is meant by an angle being bisected intemaUy and
eQctemaUyi

Prove that the internal and external bisectors of an angle are
at right angles to one another.

3. Prove that the sum of the angles formed by any number of
straight lines drawn from a point is equal to four right angles.

4. Why must every triangle have at least two a^tUe angles 1
Quote the enunciation oi the proposition from which this inference
is drawn.

5. In the enunciation The greater side of a triangle has the greater
angle opposite to it, point out what is assumed and what is to be
proved.

6. What is meant by the Proof hy Exhaustion ? Illustrate the
use of this method by naming the steps in the proof of Proposi-
tion 19.

7. What inference may be drawn respecting the triangles whose
sides measure

(i) 4 inches, 5 inches, 4 inches ;

(ii) 8 inches, 9 inches, 10 inches ;

(iii) 6 inches, 10 inches, 4 inches ?

8. Quote the enunciations of propositions which, from a hypo-
thesis relating to the sides of triangle, establish a conclusion relatmg
to the angles,

9. Quote the enunciations of propositions which, from a hypo-
thesis relating to the angles of a triangle, establish a conclusion
relating to the sides.

10. Explain why parallel straight lines must be in the same
plane,

11. Prove by means of Prop. 7 that on a given base and on the
same side of it only one equilateral triangle can be drawn.

12. In an isosceles triangle, if the equal sides are produced,
shew that the angles on the other side of the base must be obtuae«

EUCLID S ELEMENTS.

Proposition 22. Problem.

To describe a triangle having its sides equal to three given
straight lines, any two of which are together greater than the
third.

Let A, B, C be the three given straight Hnes, of which
any two are together greater than the third.

It is required to describe a triangle of which the sides shaU be
equal to A, B, C

Construction. Take a straight line DE terminated at the

point D> but unlimited towards E.
Make DF equal to A, FG equal to B, and GH equal to C. I. 3.

With centre F and radius FD, describe the circle DLK.
With centre G and radius GH, describe the circle MHK
cutting the former circle at K.

Join FK, GK.
Then shall the triangle KFG have its sides equal to the
three straight lines A, B, C.

Prool Because F is the centre of the circle DLK,

therefore FK is equal to FD : Bef. 15.

but FD is equal to A ; Constr,

therefore also FK is equal to A. Ax, 1.

Again, because G is the centre of the circle MHK,

therefore GK is equal to GH : Def, 15.

but GH is equal to C ; Constr,

therefore also GK is equal to C. Ax, 1.

And FG is equal to B. Constr,

Therefore the triangle KFG has its sides KF, FG, GK equal

respectively to the three given lines A, B, C. Q.E.F.

book i. prop. 23. 47

Proposition 23. Problem.

At a given point in a given straight line, to make an angle
equal to a given rectilineal angle.

M B

Let AB be the given straight line, and A the given point
in it, and let LCM be the given angle.

It is required to draw from A a straight line making with
AB an angle equal to the given angle DCE.

Construotion. In CL, CM take any points D and E ;

and join DE.
From AB cut off AF equal to CD. i. 3.

On AF describe the triangle FAG, having the remaining
sides AG, GF equal respectively to CE, ED. i. 22.

Then shall the angle FAG he equal to the angle DCE.

Proof. For in the triangles FAG, DCE,

(FA is equal to DC, Constr.

and AG is equal to CE ; Constr,

and the base FG is equal to the base DE: Constr.
therefore the angle FAG is equal to the angle DCE. i. 8.

That is, AG makes with AB, at the given point A, an
angle equal to the given angle DCE. q.e.f.

£jX!£iRUlS£.

On a given base describe a triangle, whose remaining sides shall
be equal to two given straight lines. Point out how the construc-
tion fails, if any one of the three given lines is greater than the sum
of the other two.

48 EUCLID'S ELEMENTS.

Proposition 24. Theorem.

If two tricmgles have two sides of the one equal to two sides
of the other, each to each, hvi the angle contained by the tivo
sides of one greater than the angle contained by the corresponding
sides of the other ; then the hose of that which has the greater
angle shall he greater than the hose of the other.

B C E _^

Let ABC, DEF be two triangles, in which
the side BA is equal to the side ED,
and the side AC is equal to the side DF,
but the angle BAC is greater than the angle EOF.

Then shaU the base BC be greater than the base EF,

Of the two sides DE, DF, let DE be that which is not
greater than the other.*

Construction. At D in the straight line ED, and on the
same side of it as DF, make the angle EDG equal to the
angle BAC. I. 23,

Make DG equal to DF or AC ; I. 3.

and join EG, GF.

Proof. Then in the triangles BAC. EDG,

BA is equal to ED, Hyp.

and AC is equal to DG, Constr,

also the contained angle BAC is equal to the

, contained angle EDG ; Constr,

therefore the triangle BAC is equal to the triangle EDG in

all respects : i. 4.

so that the base BC is equal to the base EG.

Because

BOOK I. PROP. 24. 49

Again, in the triangle FDQ,
^cause DG is equsd to DF,
therefore the angle DFG is equal to the angle DGF. i. 5,

But the angle DGF is greater than its part the angle EGF ;
therefore also the angle DFG is greater than the angle EGF ;
still more then is the angle EFG greater than the angle EGF.

And in the triangle EFG,
because the angle EFG is greater than the angle EGF,
therefore the side EG is greater than the side EF ; 1. 19.
but EG was shewn to be equal to BC ;

therefore BC is greater than EF. Q.E.D.

*The object of this step is to make the point F.fall bdow EG.
Otherwise F might fall above, upon, or below EG ; and each case
would require separate treatment. But as it is not proved that this
condition fulfils its object, this demonstration of Prop. 24 must be
considered defective.
An alternative construction and proof are given below.

Construction. At D in ED make the angle EDG equal to the
anffle BAC ; and make DG equal to DF. Join

Then, as before, it may be shewn that the
triangle EDG = the trifuigle BAC in all respects.

Now if EG passes through F, then EG is
greater than EF ; that is, BC is greater than

But if not, bisect the angle FDG by DK,
meeting EG at K. Join FK.

Proof. Then in the triangles FDK, GDK,

r FD=GD,

Because ■{ and DK is common to both,

I and the angle F D K = the angle GDK; Constr.
.'. FK=GK. I. 4.

But in the triangle EKF, the two sides EK, KF are greater than EF ;

that is, EK, KG are greater than EF.
Henee EG (or BC) is greater than EF.

H.8.BL

euclid's elements.
Proposition 25. Theorem.

If two triangles have two sides of the one equal to two sides of
the other, each to each, hut the hose of one greater than the hose
of the other ; then the angle contained by the sides of that which
has the greater hose, shall he greater than the angle contained
by the corresponding sides of the other.

Let ABC, DEF be two triangles in which

the side BA is equal to the side ED,

and the side AC is equal to the side DF,

but the base BC is greater than the base EF.

Then shall the angle BAC he greater than the angle EOF.

Proof. For if the angle BAC be not greater than the
angle EDF, it must be either equal to, or less than the
angle EDF.

But the angle BAC is not equal to the angle EDF,
for then the base BC would be equal to the base EF ; I. 4.

but it is not. Hyp,

Neither is the angle BAC less than the angle EDF,
for then the base BC would be less than the base EF ; i. 24.

but it is not. Hyp.

Therefore the angle BAC is neither equal to, nor less than

the angle EDF ;
that is, the angle BAC is greater than the angle EDF. Q.E.D.

exercise.

In a triangle ABC, the vertex A is joined to X, the middle
point of the base BC ; shew that the angle AXB is obtuse or acute,
according as AB is greater or less than AQ.

book i. prop. 26. 51

Proposition 26. Theorem.

If two triangles have two angles of the one equal to two angles
of the other, each to each, and a side of one equal to a side of
the other, these sides being either adjacent to the equal angles, or
opposite to equal angles in each; then shall the triangles be
e^wl in all respects.

Case I. When the equal sides are adjacent to the equal
angles in the two triangles.

A D

Let ABC, DEF be two triangles, in which

the angle ABC is equal to the angle DEF,

and the angle ACB is equal to the angle DFE,

and the side BC is equal to the side EF.

ITien shall the tiiangle ABC be equal to the triangle DEF m all

respects ; that is, AB shall be equal to DE, and AC to DF,

and the angle BAC shall be £qual to the angle EDF.

For if AB be not equal to DE, one must be greater than
the other. If possible, let AB be greater than DE.

Construction. From BA cut off BG equal to ED, I. 3.

and join GC.

Proof. Then in the two triangles GBC, DEF,

(GB is equal to DE, Constr.

and BC is equal to EF, Hyp.

also the contained angle GBC is equal to the
contained angle DEF ; Hyp,

therefore the triangle GBC is equal to the triangle DEF in
all respects ; i. i

so that the angle GCB is equal to the angle DFE.
But the angle ACB is equal to the angle DFE ; Hyp.
therefore also the angle GCB is equal to the angle ACB; Ax. 1.
the part equal to the whole, which is impossible.

EUCJilDS ELEMENTS.

Therefore AB is not unequal toDE ;
that is, AB is equal to DE.

Hence in the triangles ABC, DEF,

AB is equal to DE, Proved.

Because ^ *^^ ^^ ^^ ^^^ ^ ^^ ^ ^^'

also the contained angle ABC is equal to the

, contained angle DEF ; ^VP'

therefore the triangle ABC is equal to the triangle DEF in

all respects : I. 4.

so that the side AC is equal to the side DF ;

and the angle BAC is equal to the angle EDF.

Q.E.D.

Case II. When the equal sides are opposite to equal
angles in the two triangles.

H C

Let ABC, DEF be two triangles, in which

the angle ABC is equal to the angle DEF,
and the angle ACB is equal to the angle DFE,
and the side AB is equal to the side DE.

Then the triangle ABC shall he equal to the triangle DEF in all
respects ;

rmmely^ BC shaU, he eqiwX fe EF,
avid AC shall he equal to DF,
amd the angle BAC shall be equal to the a/ngle EDF.

BOOK I. PROP. 26. 53

For if BC be not equal to EF, one must be greater than
the other. If possible, let BC be greater than EF.

CoDstructioxi. From BC cut off BH equal to EF, I. 3.

and join AH.

Because

Proot Then in the triangles ABH, DEF,

AB is equal to DE, Hyp.

and BH is equal to EF, Constr.

also the contained angle ABH is equal to the

, contained angle DEF ; Hyp,

therefore the triangle ABH is equal to the triangle DEF in

all respects ; i. 4.

so that the angle AHB is equal to the angle DFE.

But the angle DFE is equal to the angle ACB ; Hyp.

therefore the angle AHB is equal to the angle ACB ; Ax. 1.

that is, an exterior angle of the triangle ACH is equal to an

interior opposite angle ; which is impossible. i. 16.

Therefore BC is not unequal to EF,
that is, BC is equal to EF.

Hence in the triangles ABC, DEF,

!AB is equal to DE, Hyp.

and BC is equal to EF ; Proved.

also the contained angle ABC is equal to the
contained angle DEF; Hyp.

therefore the triangle ABC is equal to the triangle DEF in
all respects ; i. 4.

so that the side AC is equal to the side DF,
and the angle BAC is equal to the angle EDF.

Q.E.D.

Corollary. In both cases of this Proposition it is seen
thai the triangles may be made to coincide with one another ;
and they aa-e (herefore equal in area.

54 EUCLID'S ELEMENTS,

ON THE IDENTICAL EQUALITY OF TRIANGLES.

Three cases have been already dealt with in Propositions 4, 8, and
26, the results of which may be summarized as follows :

Two triangles are equal in all respects when the following three
parts in each are severally equal :

1. Two sides, and the included angle. Prop, 4.

2. The three sides. Prop, 8, Cor.

3. (a) Two angles, and the adjacent side ; \ p^.^-, og
{h) Two angles, and a side opposite one of them. / ^*

Two triangles are not, however, necessarily equal in all respects
when any three parts of one are equal to the corresponding parts of
the other. For example

(i) When the three angles of one are
equal to the three angles of the other,
each to each, the adjoining diagram
shews that the triangles need not be
equal in all respects.

(ii) When two sides and one angle in one are equal to two sides
and one angle in the other, the
given angles being opposite to Jp

equal sides, the diagram shews
that the triangles need not be
equal in all respects.

For it will be seen that if
AB = DE, and AC=DF, and
the angle ABC = the angle
DEF, then the shorter of the given sides in the triangle DEF may
lie in either of the positions DF or DF'.

In cases (i) and (ii) a further condition must be given before we
can prove that the two triangles are identically equal.

We observe that in each of the three cases in which two triangles
have been proved equal in all respects, namely in Propositions 4, 8,
26, it is shewn that the triangles may be made to coincide with one
another ; so that they are equal in area. Euclid however restricted
himself to the use of Prop. 4, when he required to deduce the
equality in area of two triangles from the equality of certain of their
parts. This restriction is now generally abandoned.

EXERCISES ON PROPS. 12-26. 55

EXERCISES ON PROPOSITIONS 12-26.

1. If BX and CY, the bisectors of the angles at the base BC of
an isosceles triangle ABC, meet the opposite sides in X and Y, shew
that the triangles YBC, XCB are equal in all respects.

2. Shew that the perpendiculars drawn from the extremities of
the base of an isosceles triangle to the opposite sides are equal.

3. Any poiiU on the bisector of an angle is equidistant from the
arms of the angle.

4. Through O, the middle point of a straight line AB, any
straight line is drawn, and perpendiculars AX and BY are dropped
upon it from A and B : shew that AX is equal to BY.

5. If the bisector of the vertical angle of a triangle is at right
angles to the base, the triangle is isosceles.

6. The perpendicular is the shortest straight line that can he drawn
from a given point to a given straight line ; and of others, t?ujU which
is nearer to the perpendicular is less than the more remote; and two ^
and only two equaX straight lines can he drawn from the given point to
the given straight line^ one on each side of the perpendicular.

7. From two given points on the same side of a given straight line,
draw two straight lines, which shaU m^et in the given straight line,
and make equal angles with it.

Let AB be the given straight line,
and P, Q the given points.

It is required to draw from P and Q
to a point in AB, two straight lines
that shall be equally inclined to AB.

Construction. From P draw PH

Perpendicular to AB : produce PH to ^

^ making HP' equal to PH. Draw QP', meeting AB in K. Join

r K.

Then PK, QK shall be the required lines. [Supply the proof.]

8. In a given straight line find a poinfc which is equidistant
from two given intersecting straight lines. In what case is this
impossible ?

9. Through a given point draw a straight line such that the per-
pendiculars drawn to it from two given points may be equal.

In what case is this impossible ?

56 fiUCLID'S ELEMENTS.

SECTION 11.

PARALLEL STRAIGHT LINES AND PARALLELOGRAMS.

Definition. Parallel straight lines are such as, being
in the same plane, do not meet however far they are pro-
duced in both directions.

When two straight lines AB, CD are met by a third straight
line EF, eigJU angles are formed, to which
for the sake of distinction particular names
are given.

Thus in the adjoining figure,

1, 2, 7, 8 are called exterior angles,

3, 4, 5, 6 are called interior angles;

4 and 6 are said to be alternate angles ;

so also the angles 3 and 5 are alternate to
one another.

Of the angles 2 and 6, 2 is referred to as
the exterior angle, and 6 as the interior opposite angle on the
same side of EF.

2 and 6 are sometimes called corresponding angles.

So also, 1 and 5, 7 and 3, 8 and 4 are corresponding angles.

EucUd's treatment of parallel straight lines is based upon his
twelfth Axiom, which we here repeat.

Axiom 12. If a straight line cut two straight lines so
as to make the two interior angles on the same side of
it together less than two right angles, these straight lines,
being continually produced, will at length meet on that
side on which are the angles which are together less than
two right angles.

Thus in the figure given above, if the two angles 3 and 6 are
together less than two right angles, it is asserted that AB and CD
wul meet towards B and D.

This Axiom is used to establish i. 29 : some remarks upon it will
be found in a note on that Proposition.

BOOK I. PROP. 27. 57

Proposition 27. Theorem.

If a straight line, falling on two other straight lines, make
the altemate angles equal to one another, then these two straight
lines shall he parallel.

Let the straight line EF cut the two straight lines AB,
CD at Q and H, so as to make the alternate angles AGH,
GHD equal to one another.

Then shall AB and CD he parallel.

Proof. For if AB and CD be not parallel,
they will meet, if produced, either towards B and D, or

towards A and C.
If possible, let AB and CD, when produced, meet towards B

and D, at the point K.

Then KGH is a triangle, of which one side KG is produced

to A;
therefore the exterior angle AGH is greater than the interior

opposite angle GHK. i. 16.

But the angle AGH was given -equal to the angle GHK : Hyp,
hence the angles AGH and GHK are both equal and unequal ;

which is impossible.

Therefore AB and CD cannot meet when produced towards

B and D.
Similarly it may be shewn that they cannot meet towards

A and C :

therefore AB and CD are parallel.

68 euclid's elements.

Proposition 28. Theorem.

If a straight liney falling on two other straight lines, make
an exterior angle equal to the interior opposite angle on the same
side of the line ; or if it make the interior angles on the sa/me
side together equal to two right angles, then the two straight lines

shall be parallel,

Let the straight line EF cut the two straight lines AB,
CD in G and H : and

First, let the exterior angle EGB be equal to the interior
opposite angle GHD.

Then shall AB and CD be parallel.

Proot Because the angle EGB is equal to the angle GHD ;
and because the angle EGB is also equal to the vertically
opposite angle AGH ; I. 15.

therefore the angle AGH is equal to the angle GHD ;

but these are alternate angles ;
therefore AB and CD are parallel. i. 27.

Q.E.D.

Secondly, let the two interior angles BGH, GHD be to-
gether equal to two right angles.

Then shall AB and ODbe parallel,

Proot Because the angles BGH, GHD are together equal

to two right angles ; Hyp.

and because the adjacent angles BGH, AGH are also together

equal to two right angles ; I. 13.

therefore the angles BGH, AGH are together equal to the

two angles BGH, GHD.

From these equals take the common angle BGH :
then the remaining angle AGH is equal to the remaining
angle GHD : and these are alternate angles ;

therefore AB and CD are parallel. I. 27.

Q.E.D.

book i. prop. 29. 59

Proposition 29. Theorem.

If a straight line fall on two parallel straight lines, then ii
shall make the alternate angles equal to one another, and the
exterior angle eqiud to the interior opposite angle on the same
side ; and also the two interior angles on the same side equal to
two right angles.

Let the straight line EF fall on the parallel straight
lines AB, CD.

Then (i) the angle AGH shall be equal to the alternate angle
QHD;

(ii) the exterior angle EGB shall be equal to the interior
opposite angle GH D ;

(iii) the two interior angles BGH, GHD shall be together
equal to ttvo right angles,

Proot (i) For if the angle AGH be not equal to the angle

GHD, one of them must be greater than the other.
If possible, let the angle AGH be greater than the angle
GHD;

add to each the angle BGH :
then the angles AGH, BGH are together greater than the

angles BGH, GHD.
But the adjacent angles AGH, BGH are together equal to
two right angles ; i. 13.

therefore the angles BGH, GHD are together less than two

right angles ;
therefore, by Axiom 12, AB and CD meet towards B and D.
But they never meet, since they are parallel. Hyp,

Therefore the angle AGH is not unequal to the angle GHD :
that is, the angle AGH is equal to the alternate ang^le GHD.

60 EUCLID'S ELEMENTS.

(ii) Again, because the angle AGH is equal to the vertically
opposite angle EGB ; I. 15.

and because the angle AGH is equal to the angle GHD ;

Proved.

therefore the exterior angle EGB is equal to the interior
opposite angle GHD.

(iii) Lastly, the angle EGB is equal to the angle GHD ;

Proved,
add to each the angle BGH ;
then the angles EGB, BGH are together equal to the angles

BGH, GHD.
But the adjacent angles EGB, BGH are together equal to
two right angles : I. 13.

therefore also the two interior angles BGH, GHD are to-
gether equal to two right angles. Q.E.D.

EXERCISES ON PROPOSITIONS 27, 28, 29.

1. Two straight lines AB, CD bisect one another at O : shew
that the straight lines joining AC and BD are parallel. [i. 27.]

2. Straight lines which are perpendicular to the same straight line
are parallel to one another, [i. 27 or i. 28.]

3. If a straight line meet two or more parallel straight lines^ and is
perpendicular to one of them, it is also perpendicular to aUthe others,

[I. 29.]

4. If two straight lines are parallel to two other straight lines, ecich
to each, then the angles contained by the first pair are eqml respectively
to the angles contained by the second pair, [i. 29.]

BOOK I PROP. 29. 61

Note on the Twelfth Axiom.

Euclid's twelfth Axiom is unsatisfactory as the basis of a theory
of parallel straight lines. It cannot be regarded as either simple or
self-evident, ana it therefore falls short of the essential character-
istics of an axiom : nor is the difficulty entirely removed by con-
sidering it as a corollary to Proposition 17, of which it is the
converse.

Of the many substitutes which have been proposed, we need only
notice the following :

Axiom. Tvjo intersecting straight lines cannot he both parallel to a
third straight line.

This statement is known as Playfair's Axiom ; and though it is
not altogether free from objection, it is no doubt simpler and more
fundamental than that employed by Euclid, and more readily
admitted without proof.

Propositions 27 and 28 having been proved in the usual way, the
first part of Proposition 29 is then given thus.

Proposition 29. [Alternative Proof.]

If a straight line fall on two parallel straight lines, then it
shall make the alternate angles equal.

Let the straight lind EF meet the two parallel straight lines AB,
CD at G and H.

Then shaU the altemate angles AGH, GHD ^^

he equai.

For if the angle AGH is not equal to the
angle GHD:

at Q in the straight line HG make the
angle HGP equal to the angle GHD, and q
altemate to it. i. 23.

Then PG and CD are parallel. i. 27.

But AB and CD arc parallel : Syp-

therefore the two intersecting straight lines AG, PG are both parallel
to CD J

which is impossible. Playfair's Axiom,

Therefore the angle AGH is not unequal to the angle GHD ;
that is, the alternate angles AGH, GHD are equal. q.e.d.

The second and third parts of the Proposition may then be deduced
as in the text ; and Euclid's Axiom 12 follows as a Corollary.

62 EUCLID'S ELEMENTS.

Proposition 30. Theorem.

Straight lines which are pa/rallel to the same straight line are
parallel to one another,

Let the straight lines AB, CD be each parallel to the
straight line PQ.

Then shall AB and CD be parallel to one another.

Construction. Draw any straight line EF cutting AB,
CD, and PQ in the points G, H, and K.

Proot Then because AB and PQ are parallel, and EF
meets them,

therefore the angle AGK is equal to the alternate angle QKQ

L 29.
And because CD and PQ are parallel, and EF meets them,
therefore the exterior angle GHD is equal to the interior
opposite angle GKQ. I. 29.

Therefore the angle AGH is equal to the angle GHD;
and these are alternate angles ;
therefore AB and CD are parallel. I. 27.

Q.E.D.

Note. If PQ lies between AB and CD, the Proposition may be
established in a similar manner, though in this case it scarcely needs
proof ; for it is inconceivable that two straight lines, which do not
meet an intermediate straight line, should meet one another.

The truth of this Proposition may be readily deduced from
Playfair's Axiom, of which it is the converse.

For if AB and CD were not parallel, they would meet when pro-
duced. Then there would be two intersecting straight lines both
parallel to a third straight line : which is impossible.
Therefore AB apd CD never meet ; that is, they are parallel.

BOOK I. PROP. 31. 63

Proposition 31. Problem.

To draw a straight line through a given point parallel to a
given straight line.

B D C

Let A be the given point, and BC the given straight line.
It is required to draw through A a straight line parallel to BC.

Construotion. In BC take any point D ; and join AD.
At the point A in DA, make the angle DAE equal to the
angle ADC, and alternate to it, i. 23.

and produce EA to F.
Then shall EF be parallel to BC.

Proot, Because the straight line AD, meeting the two
straight lines EF, BC, makes the alternate angles EAD, ADC
equal; Constr,

therefore EF is parallel to BC ; i. 27.

and it has been drawn through the given point A.

Q.E.F.

EXERCISES.

^ 1. Any straight line drawn parallel to the base of an isosceles
triangle makes equal angles with the sides.

2. If from any point in the bisector of an angle a straight line is
drawn parallel to either arm of the angle, the triangle thus formed
is isosceles.

3. From a given point draw a straight line that shall make with
a given straight line an angle equal to a given angle.

4. From X, a point in the base BC of an isosceles triangle ABC,
a straight line is drawn at right angles to the base, cutting AB in Y,
and CA produced in Z : shew the triangle AYZ is isosceles.

5. If the straight line which bisects an exterior angle of a triangle
is parallel to the opposite side, shew that the triangle is isoscelest

64 euclid's elements^

Proposition 32. Theorem.

If a side of a triangle be produced, then the exterior angle
shall be equal to the sum of the two interior opposite angles ;
also the three interior angles of a triangle are together equal to
two right angles,

Let ABC be a triangle, and let one of its sides BC be
produced to D.

Then (i) the exterior angle ACD shall be equal to the sma of the
two interior opposite angles CAB, ABC ;

(ii) the three interior angles ABC, BCA, CAB shall be
together equal to two right angles.

Construction. Through C draw CE parallel to BA L 31.

Proof, (i) Then because BA and CE are parallel, and AC

meets them,
therefore the angle ACE is equal to the alternate angle

CAB. I. 29.

Again, because BA and CE are parallel, and BD meets them,
therefore the exterior angle ECD is equal to the interior
opposite angle ABC. I. 29.

Therefore the whole exterior angle ACD is equal to the sum
of the two interior opposite angles CAB, ABC.

(ii) Again, since the angle ACD is equal to the sum of
the angles CAB, ABC ; Proved.

to each of these equals add the angle BCA :
then the angles BCA, ACD are together equal to the three

angles BCA, CAB, ABC.
But the adjacent angles BCA, ACD are together equal to
two right angles. I. 13.

Therefore also the angles BCA, CAB, ABC are together equal
• to two right anglea Q.E.D.

BOOK I. PROP. 32. 65

From this Proposition we draw the following important
inferences.

1. If two triangles have two angles of the one equal to two angles of
the other, each to ea^h, then the third angle of the one is equal to the
third angle of the other.

2. In any right-angled triangle the two acute angles are com-
plementary,

3. In a right-angled isosceles triangle each of the equal angles is
half a right angle.

4. If one angle of a triangle is equal to the sum of the other two, the
triangle is right-angled.

6. The sum of the angles of any quadrilateral figure is equal to
four right angles.

6. Each angle of an equHaleral triangle is two-thirds of a right
angle.

EXERCISES ON PROPOSITION 32.

1. Prove that the three angles of a triangle are together equal to
two right angles,

(i) by drawing through the vertex a straight line parallel
to the base ;

(ii) by joining the vertex to any point in the base.

2. If the base of any triangle is produced both ways, shew that
the sum of the two exterior angles diminished by the vertical angle
is equal to two right angles.

3. If two straight lines are perpendicular to two other straight
lines, ea^h to ea^h, the acute angle between the first pair is equal
to the a/mte angle between the second pair.

4. Every right-angled triangle is divided into two isosceles triangles
by a straight line dravmfrom the right angle to the middle point of the
hypotenuse.

Hence the joining line is equal to half the hypotenuse,

6. Draw a straight line at right angles to a given finite straight
line from one of its extremities, without producing the given straight
line,

[Let AB be the given straight line. On AB describe any isosceles
triangle ACB. Produce BC to D, making CD equal to bC. Join
AD. Then shall AD be perpendicular to AB.]

n.S K B

66 EUCLID'S ELEMENTS.

6. Trisect a right angle.

7. The angle contained by the bisectors of the angles at the base
of an isosceles triangle is equal to an exterior angle S)rmed by pro-
ducing the base.

8. The angle contained by the bisectors of two adjacent angles of
a quadrilateral is equal to half the sum of the remaining angles.

The following theorems were added as corollaries to Proposition
32 by Robert Simson, who edited Euclid's text in 1756.

Corollary 1. All the interior angles of any rectilineal
figure, together with four right angles, are equal to twice as
rrmny right angles as the figure has sides.

Let ABODE be any rectilineal figure.

Take F, any point within it,

and join F to each of the angular points of the figure.

Then the figure is divided into as many triangles as it has
sides.

And the three angles of each triangle are together equal to
two right angles. I. 32.

Hence all the angles of all the triangles are together equal
to twice as many right angles as the figure has sides.

But all the angles of all the triangles make up all the
interior angles of the figure, together with the angles at
F, which are equal to four right angles. I. 15, Cor,

Therefore all the interior angles of the figure, together
with four right angles, are equal to twice as many right
angles as the figure has sides. Q.E.D.

BOOK I. PROP. 32. 67

Corollary 2. If the sides of a rectilineal figure^ which has
no re-entrant angle, are produced in order, then all the exterior
angles so formed are together equal to fau/r right anngles.

For at each angular point of the figure, the interior angle
and the exterior angle are together equal to two right
angles. L 13.

Therefore all the interior angles, with all the exterior
angles, are together equal to twice as many right angles
as the figure has sides.

But all the interior angles, with four right angles, are
together equal to twice as many right angles as the
figure has sides. i. 32, Cor. 1.

Therefore all the interior angles, with all the exterior
angles, are together equal to all the interior angles, with
four right angles.

Therefore the exterior angles are together equal to four
right angles. q.e.d.

EXERCISES ON SIMSON'S COROLLARIES.

[A polygon Is said to be regular when it has all its sides and all
its angles equal.]

1. Express in terms of a right angle the magnitude of each angle
of (i) a regular hexagon, (ii) a regular octagon.

2. If one side of a regular hexagon is produced, shew that the
exterior angle is equal to the angle of an equilateral triangle.

3. Prove Simson's first Corollary by joining one vertex of the
rectilineal figure to each of the other vertices.

4. Find the magnitude of each angle of a regular polygon of
n sides.

5. If the alternate sides of any polygon be produced to meet, the
sum of the included angles, together with eight right angles, will
be equal to twice as many right angles as the figure has sides.

68 EUCLID'S ELEMENTS.

Proposition 33. Theorem.

The straight lines which join tlie extremities of two equal and
parallel straight lines towards the same parts are themselves
equal and parallel.

Let AB and CD be equal and parallel straight lines ;
and let them be joined towards the same parts by the
straight lines AC and BD.

. Then shall AC and BD be equal and parallel

Construction. Join BC.

Proof. Then because AB and CD are parallel, and BC
meets them,

therefore the angle ABC is equal to the alternate angle
DCB. L 29.

Now in the triangles ABC, DCB,

(AB is equal to DC, Hyp.

and BC is common to both ;
also the angle ABC is equal to the angle
DCB ; Proved,

therefore the triangle ABC is equal to the triangle DCB in
all respects ; i. 4.

so that the base AC is equal to the base DB,
and the angle ACB equal to the angle DBC.

But these are alternate angles.
Therefore AC and BD are parallel : i. 27.

and it has been shewn that they are also equal.

Q.KD.

Definition. A Parallelogram is a four-sided figure
whose opposite sides are parallel.

BOOK I. PROP. 34. 69

Proposition 34. Theorem.

The opposite sides and angles of a parallelogram are equal
to one another, and each diagonal bisects the parallelogram.

C D

Let ACDB be a parallelogram, of which BC is a diagonal.

Then shall the opposite sides and angles of the figure be equal
to one another ; and the diagonal BC shizll bisect it

Proot Because AB and CD are parallel, and BC meets
them,

therefore the angle ABC is equal to the alternate angle
DCB; I. 29.

Again, because AC and BD are parallel, and BC meets
them,

therefore the angle ACB is equal to the alternate angle
DBC. I. 29.

Hence in the triangles ABC, DCB,

{the angle ABC is equal to the angle DCB,
and the angle ACB is equal to the angle DBC ;
also the side BC is common to both ;
therefore the triangle ABC is equal to the triangle DCB in
all respects ; I. 26.

so that AB is equal to DC, and AC to DB ;
and the angle BAC is equal to the angle CDB.

Also, because the angle ABC is equal to the angle DCB,
and the angle CBD equal to the angle BCA,
therefore the whole angle ABD is equal to the whole angle
DCA

And the triangles ABC, DCB having been proved equal in

all respects are equal in area.
Therefore the diagonal BC bisects the parallelogram ACDB.

70 EUCLID'S ELEMENTS.

EXERCISES ON PARALLELOGRAMS.

1. If one angle of a parallelogram is a right anghy oil its angles
are riglit angles.

2. If the opposite sides of a quadrilateral are equal, the figure is
a parallelogram.

3. If the opposite angles of a quadrilateral are equal, the, figure is
a parallelogram.

4 If a quadrilateral has all its sides equal and one angle a right
angle, all its angles are right angles,

5. The diagonals of a parallelogram bisect ea^h other.

6. If the dia{ionals of a quadrilateral bisect each otlier, the figure
is a parallelogram.

7. If two opposite angles of a parallelogram are bisected by the
diagonal which joins them, the figure is equilateral

8. If the diagonals of a parallelogram are equal, all its angles
are right angles.

9. In a parallelogram which is not rectangular the diagonals
are unequal.

10. Any straight line drawn through the middle point of a
diagonal of a parallelogram and terminated by a pair of opposite
sides, is bisected at that point.

11. If two parallelograms have two adjacent sides of one equal to
tujo adjacent sides of the other, ea^ch to each^ and one angle of one equal
to one angle of the other, the parallelograms are equal in all respects.

12. Two rectangles are equal if two adjacent sides of one are equal
to two adjacent sides of the other, each to each.

13. In a parallelogram the perpendiculars drawn from one pair
of opposite angles to the diagonal which joins the other pair are
equal.

14. If ABCD is a parallelogram, and X, Y respectively the middle
points of the sides AD, BC ; shew that the figure AYCX is a parallelo-
gram.

MISCELLANEOUS EXErRCISES ON SECTIONS 1. AND IL 71

MISCELLANEOUS EXERCISES ON SECTIONS I. AND II.

1. Shew that the construction in Proposition 2 may generally be
performed in eight different ways. Point out the exceptional case.

2. The bisectors of two vertically opposite angles are in the
same straight line.

3. In the figure of Proposition 16, if AF is joined, shew

(i) that AF is equal to BC ;

(ii) that the triangle ABC is equal to the triangle CFA in all
respects.

4. ABC is a triangle right-angled at B, and BC is produced to'
D : shew that the angle ACD is obtuse.

5. Shew that in any regular polygon of n sides each angle

contains M right angles.
n

6. The angle contained by the bisectors of the angles at the
base of any triangle is equal to the vertical angle together with half
the sum of the base angles.

7. The angle contained by the bisectors of two exterior angles
of any triangle is equal to half the sum of the two corresponding
interior angles.

8. If perpendiculars are drawn to two intersecting straight lines
from any point between them, shew that the bisector of the angle
between the perpendiculars is parallel to (or coincident with) the
bisector of the angle between the given straight lines.

9. If two points P, Q be taken in the equal sides of an isosceles
triangle ABC, so that BP is equal to CQ, shew that PQ is parallel
toBC.

10. ABC and DEF are two triangles, such that AB, BC are equal
and parallel to DE, EF, each to each ; shew that AC is equal and
paraUel to DF.

11. Prove the second Corollary to Prop. 32 by drawing through
any angular point lines parallel to all the sides.

12. If two sides of a quadrilateral are parallel, and the remaining
two sides equal but not parallel, shew that the opposite angles are
supplementary ; also that the diagonals are equal

72 Euclid's elemjbnts.

SECTION m.

THE AREAS OF PARALLELOGRAMS AND TRIANGLES.

Hitherto when two figures have been said to be equcUf it has
been implied that they are identicaUy equal, that is, equal iu all
respects.

But figures may be equal in area without being equal in all
respects, that is, without having the same shape.

The present section deals with parallelograms and triangles which
are equal in area but not necessarily identically equal.

[The ultimate test of equality, as we have already seen, is afforded
by Axiom 8, which asserts that magnitudes which may he made to
coincide with one another are equal. Now figures which are not equal
in all respects, cannot be made to coincide without first undergoing
some change of form : hence the method of direct superposition is
unsuited to the purposes of the present section.

We shall see however from Euclid's proof of Proposition 36, that
two figures which are not identically equal, may nevertheless be so
related to a third figure, that it is possible to infer the equality of
their areas. ]

Definitions.

1. Tbe Altitude of a parallelogram with reference to a
given side as base, is tbe perpendicular distance between
the base and the opposite side.

2. The Altitude of a triangle with reference to a given
side as base, is the perpendicular distance of the opposite
vertex from the base.

[From this point the following symbols will be introduced into
the text :

= for is equal to ; .*. for therefore.

If it is thought desirable to shorten written work by the use of
symbols and abbreviations, it is strongly recommended that only
some well recognized system should be aUowed, such, for example,
as that given on page 11.]

BOOK L PROP. 35. 73

Proposition 35. Theorem.

Parallelograms on the same base, and between the same
parallels, are equal in area.

A DE F A DE F A E D F

B C B C B C

Let the parallelograms ABCD, EBCF be on the same
base BC, and between the same parallels BC, AF.

Then shall the parallelogram ABCD be equal in area to the
parallelogram EBCF.

Case I. If the sides AD, EF, opposite to the base BC,
are terminated at the same point D :

then each of the parallelograms ABCD, EBCF is double of

the triangle BDC ; I. 34.

.'. they are equal to one another. Ax, 6.

Case II. But if the sides AD, EF are not terminated at
the same point :

then because ABCD is a parallelogram,
/. the side AD = the opposite side BC; I. 34.

similarly EF = BC ;

.-. AD = EF. Ax. 1.

.*. the whole, or remainder, EA = the whole, or remainder, FD.

Then in the triangles FDC, EAB,

(FD = EA, Proved.

and the side DC = the opposite side AB, I. 34.
also the exterior angle FDC = the interior opposite
angle EAB, I. 29.

.*. the triangle FDC = the triangle EAB. I. 4.

From the whole figure ABCF take the triangle FDC;
and from the same figure take the equal triangle EAB ;

then the remainders are equal. Ax. 3.

Therefore the parallelogram ABCD is equal to the parallelo-
gram EBCF.

74 euclid's elements.

Proposition 36. Theorem.

Parallelograms on equal hoses, and between the same parallels^
are equal in area.

Let ABCD, EFGH be parallelograms on equal bases BC,
FG, and between the same parallels AH, BG.

Then shall the 'parallelogram ABCD he equal to the parallelo-
gram EFGH.

Oonstaruction. Join BE, CH.

Proot Then because BC = FG ; Hyp,

and the side FG = the opposite side EH ; I. 34.

.-. BC=EH: Ax.\.

and BC is parallel to EH ; Hyp.

.*. BE and CH are also equal and parallel. I. 33.

Therefore EBCH is a parallelogram. Def. 36.

Now the parallelograms ABCD, EBCH are on the same
base BC, and between the same parallels BC, AH ;
.". the parallelogram ABCD = the parallelogram EBCH. I. 35.

Also the parallelograms EFGH, EBCH are on the same
base EH, and between the same parallels EH, BG ;
.". the parallelogram EFGH = the parallelogram EBCH. L 35.

Therefore the parallelogram ABCD is equal to the parallelo-
gram EFGH. Ax. 1.

Q.E.D.

From the last two Propositions we infer that :

(i) A parallelogram is equal in area to a rectangle of equal
base and equal altitude.

(ii) Pa/rallelograms on equal bases and of equal altitudes a/re
equal in area.

BOOK I. PROP. 37, 75

Proposition 37. Theorem.

Triangles on the same base, and between the same parallels,
are equal in area.

Let the triangles ABC, DBC be upon the same base BC,
and between the same parallels BC, AD.

Then shall the triangle ABC be equal to tJie triangle DBC.

Ck>n8truction. Through B draw BE parallel to CA, to
meet DA produced in E ; i. 31.

through draw CF parallel to BD, to meet AD produced in F.

Proot Then, by construction, each of the figures EBCA,
DBCF is a parallelogram. Def, 36.

And since they are on the same base BC, and between the

same parallels BC, EF ;
.'. the parallelogram EBCA = the parallelogram DBCF. l. 35.

Now the diagonal AB bisects EBCA ; I. 34.

.*. the triangle ABC is half the parallelogram EBCA.

And the diagonal DC bisects DBCF ; I. 34.

.*. the triangle DBC is half the parallelogram DBCF.

And the halves of equal things are equal. Ax, 7,
Therefore the triangle ABC is equal to the triangle DBC.

Q.KD.

[For Exercises see page 79.]

76 EUCLID*S ELEMENTS.

Proposition 38. Theorem.

Triangles on equal bases, amd between the same parallels, are
equal in area.

Let the triangles ABC, DEF be on equal bases BC,
and between the same parallels BF, AD.

Then shall the triangle ABC be equal to the triangle DEF.

Construction. Through B draw BG parallel to CA, to
meet DA produced in G; I. 31.

through F draw FH parallel to ED, to meet AD produced in H.

Proot Then, by construction, each of the figures GBCA,
DEFH is a parallelogram. Def, 36.

And since they are on equal bases BC, EF, and between the

same parallels BF, GH;
/. the parallelogram GBCA = the parallelogram DEFH. I. 36.

Now the diagonal AB bisects GBCA ; I. 34.

.-. the triangle ABC is half the parallelogram GBCA.

And the diagonal DF bisects DEFH ; I. 34.

.'. the triangle DEF is half the parallelogram DEFH.

And the halves of equal things are equal. Ax. 7.
Therefore the triangle ABC is equal to the triangle DEF.

Q.E.D.

From this Proposition we infer that :

(i) Triangles on equal bases and of equal altitude are equal
in area,

(ii) Of two triangles of the same altitude, that is the greater
which has the greater base; and of two triangles on the same base,
or on equal bases, that is the greater which has the greater' altitude.

BOOK I. PROP. 39, 77

Proposition 39. Theorem.

Equal triangles on the same base, and on the same side of it,
are between the same parallels.

Let the triangles ABC, DBC which stand on the same
base BC, and on the same side of it be equal in area.

Hien shall the triangles ABC, DBC be between the same parallels;
that is, if AD be joined, AD shall be parallel to BC.

Construction. For if AD be not parallel to BC,

if possible, through A draw AE parallel to BC, I. 31.
meeting BD, or BD produced, in E.
Join EC.

Proot Now the triangles ABC, EBC are on the same
base BC, and between the same parallels BC, AE ;

.-. the triangle ABC = the triangle EBC. I. 37.

But the triangle ABC = the triangle DBC; Hyp,
.*. the triangle DBC = the triangle EBC;
that is, the whole is equal to a part ; which is impossible.

.*. AE is not parallel to BC.

Similarly it can be shewn that no other straight line
through A, except AD, is parallel to BC.

Therefore AD is parallel to BC.

Q.E.D.

From this Proposition it follows that :

Eqiuil triangles on the same base have equal altitudes,

[For Exercises see page 79.]

78 EUCLID'S ELEMENTS.

Proposition 40. Theorem.

Eqmt triangles^ on equal bases in the same straight line^ and
on the same side of it, are between the same parallels.

Let the triangles ABC, DEF which stand on equal bases
BC, EF, in the same straight line BF, and on the same side
of it, be equal in area.

Then shall the triangles ABC, DEF be between the sams parallels 'y
that is, if AD be joined, AD shall be parallel to BF.

Construction. For if AD be not parallel to BF,

if possible, through A draw AG parallel to BF, L 31.
meeting ED, or ED produced, in G.
Join GF.

Proof. Now the triangles ABC, GEF are on equal bases
BC, EF, and between the same parallels BF, AG ;

.-. the triangle ABC = the triangle GEF. L 38.

But the triangle ABC = the triangle DEF : Hi^.
.-. the triangle DEF = the triangle GEF:
that is, the whole is equal to a part ; which is impossible.

.-. AG is not parallel to BF.

Similarly it can be shewn that no other straight line
through A, except AD, is parallel to BF.

Therefore AD is parallel to BF.

Q.E.D.

From this Proposition it follows that :

(i) Equal triangles on equal bases have equal altitudes.
(ii) Equal triangles of equal altitudes have equal bases.

EXERCISES ON PROPS. 37-40. 79

EXERCISES ON PROPOSITIONS 37-40.

Definition. Each of the three straight lines which join
the angular points of a triangle to the middle points of
the opposite sides is called a Median of the triangle.

ON Prop. 37.

1. If, in the figure of Prop. 37, AC and BD intersect in K, shew
that

(i) the triangles AKB, DKC are equal in area. •
(ii) the quadrilaterals EBKA, FCKD are equal.

2. In the figure of i. 16, shew that the triangles ABC, FBC are
equal in area.

3. On the base of a given triangle construct a second triangle,
equal in area to the first, and having its vertex in a given straight
line.

4. Describe an isosceles triangle equal in area to a given triangle
and standing on the same base.

ON Pbop. 38.

5. A triangle is divided by each of its medians into two parts of
equal area.

6. A parallelogram is divided by its diagonals into four triangles
of equal area.

7. ABC is a triangle, and its base BC is bisected at X ; if Y
be any point in the median AX, shew that the triangles ABY, ACY
are equal in area.

8. In AC, a diagonal of the parallelogram A BCD, any point X is
taken, and XB, XD are drawn : shew that the triangle BAa is equal
to the triangle DAX.

9. If two triangles h£U(re two sides of one respectively equal to
two sides of the other, and the angles contained by those sides
supplementary f the triangles are equal in area.

ON Prop. 39.

10. The straight line lohich joins the middle points of two sides of
a triangle is parallel to the third side.

11. Iftxoo straight lines AB, CD intersect in O, so that the triangle
AOC is equal to the triangle DOB, shew that AD and CB are parallel.

ON Prop. 40.

12. Deduce Prop. 40 from Prop. 39 by joining AE, AF in the
figure of page 78.

80 EUCLID'S ELEMENTS.

Proposition 41. Theorem.

If a paralldoffram and a triangle be on the same base and
between the same parallels^ the parallelogram shall be dovble of
the triangle.

Let the parallelogram ABCD, and the triangle EBC be
upon the same base BC, and between the same parallels
BC, AE.

Then shall the parallelogram ABCD be double of the triangle
EBC.

Construction. Join AC.

Proof. Now the triangles ABC, EBC are on the same
base BC, and between the same parallels BC, AE ;

.'. the triangle ABC = the triangle EBC. I. 37.

And since the diagonal AC bisects ABCD ; I. 34.
.". the parallelogram ABCD is double of the triangle ABC.

Therefore the parallelogram ABCD is also double of the
triangle EBC. Q.E.D.

EXERCISES. •

1. ABCD is a parallelogram, and X, Y are the middle points of
the sides AD, BC ; if Z is any point in XY, or XY produced, shew
that the triangle AZB is one quarter of the parallelogram ABCD.

2. Describe a right-angled isosceles triangle equal to a given
square.

3. If ABCD is a parallelogram, and X, Y any points in DC and AD
respectively : shew that the triangles AXB, BYC are equal in area.

4. ABCD is a parallelogram^ and P is any point within it ; shew
that the sum of the triangles PAB, PCD is equal to half the paral-
lelogram.

BOOK I. PROP. 42. 81

Proposition 42. Problem.

To describe a parallelogram that shall be equal to a given
triangle, and have one of its angles equal to a given angle.

Let ABC be the given triangle, and D the given angle.

It is rehired to describe a parallelogram equal to ABC, and
having one of its angles equal to D.

Construction. Bisect BC at E. I. 10.

At E in CE, make the angle CEF equal to D ; I. 23.
through A draw AFG parallel to EC ; I. 31.

and through C draw CG parallel to EF.
Then FECG mall be the parallelogram required.

Join AE.

Proof. Now the triangles ABE, AEC are on equal bases
BE, EC, and between the same parallels ;

.'. the triangle ABE = the triangle AEC; I. 38.
.-. the triangle ABC is double of the triangle AEC.

But FECG is a parallelogram by construction ; Def 36
and it is double of the triangle AEC,
being on the same base EC, and between the same parallels
EC and AG. I. 41,

Therefore the parallelogram FECG is equal to the triangle

ABC;
and it has one of its angles CEF equal to the given angle D.

Q.E.F.

EXERCISES.

1. Describe a parallelogram equal to a given square standing on
the same base, and having an angle equal to half a right angle.

2. Describe a rhombus equal to a given parallelogram and stand-
ing on the same base. When does the construction iaAVl

EUCLID'S ELEMENTS.

Definition. If in the diagonal of a parallelogram any
point is taken, and straight lines are drawn through it
parallel to the sides of the parallelogram ; then of the four
parallelograms into which the whole figure is divided, the
two through which the diagonal passes are called Paral-
lelograms about that diagonal, and the other two, which
with these make up the whole figure, are called the
complements of the parallelograms about the diagonal.

B G

Thus in the figure given above, AEKH, KGCF are parallelograms
about the diagonal AC ; and the shaded figures HKrD, EBGrC are
the complements of those parallelograms.

Note. A parallelogram is often named by two letters only, these
being placed at opposite angular points.

BOOK L PROP. 43. 83

Proposition 43. Theorem.

Hie complements of the parallelograms ahovi the diagonal of
any parallelog^'am, are equal to one another.

Let ABCD be a parallelogram, and KD, KB the comple-
ments of the parallelograms EH, GF about the diagonal AC.

Then shall the complement BK be equal to the complement KD.

Proof. Because EH is a parallelogram, and AK its diagonal,
.-. the triangle AEK = the triangle AHK. I. 34.

Similarly the triangle KGC = the triangle KFC.
Hence the triangles AEK, KGC are together equal to the
triangles AHK, KFC.

But since the diagonal AC bisects the parallelogram ABCD ;
.'. the whole triangle ABC = the whole triangle ADC. I. 34.

Therefore the remainder, the complement BK, is equal to the
remainder, the complement KD. Q.E.D.

EXERCISES.

In the figure of Prop. 43, prove that

(i) The parallelogram ED is equal to the parallelogram BH.

(ii) If KB, KD are joined, the triangle AKB is equal to the
triangle AKD,

84 EUCLID'S ELEMENTS.

Proposition 44. Problem.

To a given straight line to apply a parallelogram which shall
be equal to a given triangle, and have one of its angles equal to
a given angle.

Let AB be the given straight line, C the given triangle,
and D the given angle.

It is required to apply to the straight line AB a parallelogram
equal to the triangle C, and having an angle equal to the angle D.

Construction. On AB produced describe a parallelogram
BEFQ equal to the triangle C, and having the angle EBG
equal to the angle D. i. 22 and i. 42*.

Through A draw AH parallel to BG or EF, to meet FG pro-
duced in H. I. 31.

Join HB.

Then because AH and EF are parallel, and HF meets them,
.*. the angles AHF, HFE together = two right angles. L 29.

Hence the angles BHF, HFE are together less than two
right angles ;

•*. H B and FE will meet if produced towards B and E. Ax, 1 2.

Produce HB and FE to meet at K.

Through K draw KL parallel to EA or FH ; I. 31.
and produce HA, GB to meet KL in the points L and M.

Then shall BL be the parallelogram required.

BOOK I. PROP. 44. 85

Proof. Now FHLK is a parallelogram, Constr,

and LB, BF are the complements of the parallelograms
about the diagonal HK :

.*. the complement LB = the complement BF. I. 43.

But the triangle C = the figure BF ; Constr.

.*. the figure LB = the triangle C.

Again the angle ABM = the vertically opposite angle QBE ;
also the angle D = the angle QBE ; Constr.

.*. the angle ABM = the angle D.

Therefore the parallelogram LB, which is applied to the
straight line AB, is equal to the triangle C, and has the
angle ABM equal to the angle D. Q.E.F.

♦ This step of the construction is effected by first describing on
AB produced a triangle whose sides are respectively equal to those of
the triangle C (i. 22) ; and by then making a parallelogram equal to
the triangle so drawn, and having an angle equal to D (i. 42).

QUESTIONS FOR REVISION.

1. Quote Euclid's Twelfth Axiom. What objections have been
raised to it, and what substitute for it has been suggested ?

2. Which of Euclid's Propositions, dealing with parallel straight
lines, depends on Axiom 12 ? Furnish an alternative prool.

3. Straight lines which are parallel to the same straight line are
parallel to one another [Prop. 30]. Deduce this from Playfair's
Axiom.

4. Define a parallelogram, an altitude of a triangle, a median of
a triangle, parallelograms about the diagonal of a parallelogram.

5. What is meant by superposition 1 On what Axiom does this
method depend ? Give instances of figures which are equal in area^
but which cannot be superposed.

6. In fig. 2 of Prop. 35 shew how one parallelogram may be cut
into pieces, which, when fitted together in other positions, make up
the other parallelogram.

EUCLID S BLBMKNTS,

Proposition 45. Problem.

To describe a paTollelogram equal to a given, TeclUineal figure,
o/nd having an angle equal to a given a/ngle.

Let ABCD be the given rectilineal figure, and E the
given angle.

// is required to desa-ibe a parallelogram eqml to ABCD,
and having an, angle equal to E.
Suppose the given rectilineal figure to be a quadrilateraL

Construction. Join BD.

Describe the parallelogram FH equal to the triangle ABD,

and having the angle FKH equal to the angle E. i. 42.

To OH apply the parallelogram GM, equal to the triangle

OBC, and having the angle GHM equal to E. L 44.

Then shall FKML be the parallelogram required.

Proof. Because each of the angles GH M, FKH = the angle E ;

.-. the angle FKH = the angle GHM.

To each of these equals add the angle OHK ;

then the angles FKH, QHK together = the angles GHM, GHK.

But since FK, GH are parallel, and KH meets them ;
.-. the angles FKH, GHK together^two right angles; I. 29.
.'. also the angles GHM, QHK together = two right angles;
.-. KH, HM are in the same straight lins.' I. 14.

BOOK I. PROP. 45. 87

Again, because KM, FG are parallel, and HG meets them,
.*. the angle MHG = the alternate angle HGF. L 29.

To each of these equals add the angle HGL ;
then the angles MHG, HGL together = the angles HGF, HGL.

But because HM, GL are parallel, and HG meets them,
.•• the angles MHG, HGL together = two right angles : I. 29.
.•. also the angles HGF, HGL together = two right angles:
.'. FG, GL are in the same straight line. i. 14,

And because KF and ML are each parallel to HG, Constr.

therefore KF is parallel to ML; i. 30.

and KM, FL are parallel ; Constr,

,', FKML is a parallelogram. Def, 36.

Again, because the parallelogram FH = the triangle ABD,
and the parallelogram GM =the triangle DBC ; Constr.
5% the whole parallelogram FKML = the whole figure ABCD ;
and it has the angle FKM equal to the angle E.

By a series of similar steps, a parallelogram may be
constructed equal to a rectilineal figure of more than four
sides. Q.E.F.

88 buclid's elements.

The following Problem is important, and furnishes a useful appli-
cation of the principles of the foregoing propositions.

ADDITIONAL PROBLEM.
To describe a triangle equal in area to a given quadrilateral.

Let ABCD be the given quadrilateral.
It is required to describe a triangle eqvjaZ to ABCD in area^

Construction. Join BD.

Through C draw CX parallel to BD, meeting AD produced in X.

Join BX.
Then XAB shall be the required triangle.

Proof. Now the triangles XDB, CDB are on the same base DB
and between the same paraUels DB, XC ;

.*. the triangle XDB = the triangle CDB in area. I. 37.

To each of these equals add the triangle ADB ;
> then the triangle XAB = the figure ABCD.

iiiJ'k.JliXvC'IbjIi.

Construct a rectilineal figure equal to a given rectilineal figure,
and having fewer sides by one than the given figure.

Hence shew how to construct a triangle equal to a given recti-
lineal figure;

BOOK I. PROP. 46.

Proposition 46. Problem.
To describe a square on a given straight line.

A B

Let AB be the given straight line.
It is required to describe a square on AB.

Constr. From A draw AC at right angles to AB ; I. 11.
and make AD equal to AB. I. 3.

Through D draw DE parallel to AB ; I. 31.

and through B draw BE parallel to AD, meeting DE in E.

Then shall ADEB be a square,

Prool For, by construction, ADEB is a parallelogram :

.-. AB=DE, and AD = BE. I. 34.

But AD = AB ; Constr.

,'. the four straight lines AB, AD, DE, EB are all equal ;
that is, the figure ADEB is equilateral.

Again, since AB, DE are parallel, and AD meets them,
.'. the angles BAD, ADE together = two right angles ; i. 29.
but the angle BAD is a right angle ; Constr,

. *. also the angle ADE is a right angle.

And the opposite angles of a parallelogram are equal ; l. 34.
.*. each of the angles DEB, EBA is a right angle :

that is the figure ADEB is rectangular.
Hence it is a square, and it is described on AB.

Q.E.F.

Corollary. If one angle of a parallelogram is a right
angle^ all its angles are right angles.

EUCLID'S ELEMENTS.

Proposition 47. Theorem.

In a right-angled triangle the square described on the
hypotenuse is equal to the sum of the squares described on the
other two sides.

Let ABC be a right-angled triangle, having the angle
BAC a right angle.

Hi&n shall the square described on the hypotenuse BC be
equal to the sum of the squares described on BA, AC.

Construction. On BC describe the square BDEC ; I. 46.
and on BA, AC describe the squares BAGF, ACKH.

Through A draw AL parallel to BD or CE ; I. 31.
and join AD, FC.

Proof Then because each of the angles BAC, BAG is a
right angle,

.-. CA and AG are in the same straight line. L 14.

Now the angle CBD = the angle FBA,

for each of them is a right angle.

Add to each the angle ABC :

then the whole angle ABD = the whole angle FBO.

BOOK I. PROP. 47. 91

Then in the triangles ABD, FBC,

I AB = FB,

Because < and BD = BC,

[also the angle ABD = the angle FBC ; Proved.

.-. the triangle ABD = the triangle FBC. I. 4.

Now the parallelogram BL is double of the triangle ABD,
being on the same base BD, and between the same parallels
BD, aL I. 41.

And the square GB is double of the triangle FBC, being
on the same base FB, and between the same parallels
FB, GC. I. 41.

But doubles of equals are equal : Ax, 6.

therefore the parallelogram BL = the square GB.

Similarly, by joining AE, BK it can be shewn that the

parallelogram CL = the square CH.

Therefore the whole square BE = the sum of the squares
GB, HC :

that is, the square described on the hypotenuse BC is equal
to the sum of the squares described on the two sides
BA, AC. Q.E.D.

Note. It is not necessary to the proof of this Proposition that
the three squares should be described external to the triangle ABC ;
and since each square may be drawn either towards or away from the
triangle, it may be shewn that there are 2x2x2, or eighty possible
constructions.

Obs, The following properties of a square, though not
formally enunciated by Euclid, are employed in subsequent
proofs. [See I. 48.]

(i) The squares on equal straight lines are equal.
(ii) Equal squares stand upon equal straight lines.

92 EUCLID'S ELEMENTS.

EXERCISES ON PROPOSITION 47.

1. In the figure of this Proposition, shew that

(i) If BG, CH are joined, these straight lines are parallel ;

(ii) The points F, A, K are in one straight line ;

(iii) FC and AD are at right angles to one another ;

(iv) If GH, KE, FD are joined, the triangle GAH is equal
to the given triangle in all respects ; and the triangles
FBD, KCE are each equal in area to the triangle AoC.

[See Ex. 9, p. 79.]

2. On the sides AS, AC of any triangle ABC, squares ABFG,
ACKH are described both toward the triangle, or both on the side
remote from it : shew that the straight lines BH and CG are equal.

3. On the sides of any triangle ABC, equilateral triangles BCX,
CAY, ABZ are described, all externally, or all towards the triangle :
shew that AX, BY, CZ are all equal.

4. The square described on the dia^gonal of a given square, is
double of the given square,

5. ABC is an equilateral triangle, and AX is the perpendicular
drawn from A to BC ; sliew that the square on AX is three times the
square on BX.

6. Describe a square equal to the sum of two given squares.

7. From the vertex A of a triangle ABC, AX is drawn perpendi-
cular to the base : shew that the diflference of the squares on the
sides AB and AC, is equal to the difference of the squares on BX and
CX, the segments of the base.

8. If from any point O within a triangle ABC, perpendiculars
OX, OY, OZ are drawn to the sides BC, CA, AB respectively : shew
that the sum of the squares on the segments AZ, BX, CY is equal to
the sum of the squares on the segments AY, CX, BZ.

9. ABC is a triangle right-angled at A ; and the sides AB, AC
are intersected by a straight line PQ, and BQ, PC are joined.
Prove that the sum of the squares on BQ, PC is equal to the sum
of the squares on BC, PQ.

10. In a right-angled triangle four times the sum of the squares
on the two medians drawn from the acute angles is equal to five
times the square on the hypotenuse.

ON BOOK I. PROP. 47.

NOTES ON PROPOSITION 47.

It is believed that Proposition 47 is due to Pythagoras, a Greek
philosopher and mathematician, who lived about two centuries before
Euclid.

Many experimental proofs of this theorem have been given by
means of actual dissection : that is to say, it has been shewn how the
squares on the sides containing the right angle may be cut up into
pieces which, when fitted together in other positions, exactly make
up the square on the hypotenuse. Two of these methods of dissec-
tion are given below.

I. In the adjoining diagram ABC is the given right-angled
triangle, and the fibres AF, HK
are the squares on AB, AC, placed
side by side.

FD is made equal to EH or AC ;

and the two squares AF, HK are cut
along the lines ED, DB.

Then it will be found that the
triangle EHD may be placed so as
to fill up the space CAB ; and the
triangle BFD may be made to fill
the space CKE.

Hence the two squares AF, HK
may be fitted together so as to
form the single figure CBDE, which
will be found to be a perfect square, namely the square on the
hypotenuse BC.

n. In the figure of i. 47, let DB F
and EC be produced to meet FG and
AH in L ana N respectively ; and let
LM be drawn parallel to BC.

Then it will be found that the
several parts of the two squares FA,
AK can be fitted together (in the places
bearing corresponding numbers) so as
exactly to fill up the square DC.

94 EUCUD'S ELBafENTS.

Proposition 48. Theorem.

If the square described on one side of a triangle be equal to
the sum of the squares described on the other tvx) sides, then the
a/ngle contained by these two sides shall be a right an^le.

Let ABC be a triangle ; and let the square described on
BC be equal to the sum of the squares described on BA, AC.

Then shall the angle BAC be a right angle.

Construction. From A draw AD at right angles to AC; I. 11.

and make AD equal to AB. L 3.

Join DC.

Proot Then, because AD = AB, Constr,

.*. the square on AD = the square on AB.
To each of these add the square on CA ;
then the sum of the squares on CA, AD = the sum of the
squares on CA, AB.

But, because the angle DAC is a right angle, Constr.
.*. the square on DC = the sum of the squares on CA, AD. 1. 47.
And, by hypothesis, the square on BC=the sum of the
squares on CA, AB ;

.'. the square on DC = the square on BC :
.*. also the side DC = the side BC.

Then in the triangles DAC, BAC,

f DA=BA, Constr.
Because < and AC is common to both ;

[also the third side DC = the third side BC; Proved.

.-. the angle DAC = the angle BAC. I. 8.

But DAC is a right angle. Constr.

Therefore also BAC is a right angle. Q.E.D.

THEOREMS AND EXAMPLES ON BOOK I.

INTEODUCTOEY.

HINTS TOWARDS THE SOLUTION OF GEOMETRICAL EXERCISES.

ANALYSIS. SYNTHESIS.

It is commonly found that exercises in Pure Geometry present to
a beginner far more difficulty than examples in any other branch of
Elementary Mathematics. This seems to be due to the following
causes :

(i) The variety of such exercises is practically unlimited ; and
it is impossible to lay down for their treatment any definite methods,
such for example as the rules of Elementary Arithmetic and Algebra.

(ii) The arrangement of Euclid's Propositions, though perhaps
the most convincing of all forms of argument, affords in most 'cases
little clue as to the way in which the proof or construction was
discovered.

Euclid's propositions are arranged synthetically : that is to say,
starting from the hjrpothesis or data, they first give a construction
in accordance with postulates, and problems already solved ; then
by successive steps based on known theorems, they prove what was
required in the enunciation.

Thus Geometrical Synthesis is a building up of known results, in
order to obtain a new result.

But as this is not the way in which constructions or proofs are
usually discovered, we draw the student's attention to the following
hints.

Begin by assv/ming the result it is desired to establish ; then by
working backwards, trace the consequences of the assumption, and
try to ascertain its dependence on some simpler theorem which is
already known to be true, or on some condition which suggests the
necessary construction. If this attempt is successful, the steps of
the argument may in general be re-arranged in reverse order, and
the construction and proof presented in a synthetic form.

This unravelling of a proposition in order to trace it back to
some earlier principle on which it depends, is called geometrical
analsrsis : it is the natural way of attacking many theorems, and it
is especially useful in solving problems.

Although the above directions do not amount to a method, they
often furnish a mode of searching for a suggestion. Geometrical
Analysis however can only be used with success when a thorough
grasp of the chief propositions of Euclid has been gained.

Euclid's elements.

The practical application of the foregoing hints is illustrated by
the following examples.

1. Construct an isosceles triangle having given the hase^ and the
sum of one of the equal sides and the perpendicular drawn from the
vertex to the base.

> I

Let AB be the given base, and K the sum of one side and the
perpendicular drawn from the vertex to the base.

Analysis. Suppose ABC to he the required triangle.

From C draw CX perpendicular to AB :

then AB is bisected at X. i. 26.

Now if we produce XC to H, making XH equal to K,
it follows that CH =CA ;
and if AH is joined,
we notice that the angle CAH = the angle CHA. i. 5.

Now the straight lines XH and AH can be drawn before the
position of C is knoivn ;

Hence we have the following construction, which we arrange
synthetically.

Synthesis. Bisect AB at X :

from X draw XH perpendicular to AB, making XH equal to K.

Join AH.
At the point A in HA, make the angle HAC equal to the angle AHX.

Join CB.
Then ACB shall he the triangle required.

First the triangle is isosceles, for AC = BC.
Again, since the angle HAC = the angle AHC,

/. HC = AC.
To each add CX ;
then the sum of AC, CX = the sum of HC, CX

= HX.
That is, the sum of AC, CX = K. Q.B.F.

I. 4.

Gonstr.
I. 6.

THEOREMS AND EXAMPLES ON BOOK I. 97

2. To divide a given straight line so that the square on one part
may be double of the square on the other.

Let AB be the given straight line.

Analysis. Suppose AB to he divided as required at X : that is,
suppose the sqtiare on AX to be double of the square oti XB.

Now we remember that in an isosceles right-angled triangle, the
square on the hypotenuse is double of the square on either of the
equal sides.

This suggests to us to draw BC perpendicular to A B, to make BC
equal to BX, and to join XC.

Then the square on XC is double of the square on XB ; i. 47.

XC = AX.
Hence when we join AC, we notice that

the angle XAC = the angle XCA. i. 5.

Thus the exterior angle CXB is double of the angle XAC. i. 32.

But the angle CXB is half of a right angle : i. 32.

.*. t?ie angle aAC is one-fourth of a right angle.

This supplies the clue to the following construction : —

Synthesis. From B draw BD perpendicular to AB ;
and from A draw AC, making BAC one-fourth of a right angle.
From C, the intersection of AC and BD, draw CX, making the angle
ACX equal to the angle BAC. i. 23.

Then AB shall be divided ow required at X.

For since the angle XCA = the angle XAC,

.'. XA = XC. I. 6.

And because the angle BXC=the sum of the angles BAC, ACX, i. 32.
.'. the angle BXC is half a right angle.

And the angle at B is a right angle ;
.'. the angle BCX is half a right angle ; i. 32.

.'. the angle BXC = the angle BCa ;
.-. BX = BC.
Hence the square on XC is double of the square on XB : i. 47.
that is, the square on AX is double of the square on XB. q.b.f.
H.S.B. Q

98 EUCLID'S ELEMENTS.

L ON THE IDENTICAL EQUALITY OF TRIANGLES.

See Propositions 4, 8, 26.

1. If in a triangle the perpendicular from the vertex on the base
bisects the base, then the triangle is isosceles.

2. If the bisector of the vertical angle of a triangle is also per-
pendicular to the base, the triangle is isosceles.

3. If the bisector of the vertical angle of a triangle also bisects
the base, the triangle is isosceles.

[Produce the bisector, and complete the construction after the
manner of i. 16.]

4. If in a triangle a pair of straight lines drawn from the ex-
tremities of the base, making equal angles with the remaining sides,
are equal, the triangle is isosceles.

5. If in a triangle the perpendiculars drawn from the extremities
of the base to the opposite sides are equal, the triangle is isosceles.

6. Two triangles ABC, ABD on the same base AB, and on opposite
sides of it, are such that AC is equal to AD, and BC is equal to BD :
shew that the line joining the points C and D is perpendicular to AB.

7. If from the extremities of the base of an isosceles triangle
perpendiculars are drawn to the opposite sides, shew that the
straight line joining the vertex to the intersection of these per-
pendiculars bisects the vertical angle.

8. ABC is a triangle in which the vertical angle BAC is bisected
by the straight line AX : from B draw BD perpendicular to AX, and
produce it to meet AC, or AC produced, in t ; then shew that BD is
equal to DE.

9. In a quadrilateral ABCD, AB is equal to AD, and BC is equal
to DC : shew that the diagonal AC bisects each of the angles which
it joins.

10. In a quadrilateral ABCD the opposite sides AD, BC are
equal, and also the diagonals AC, BD are eaual : if AC and BD inter-t
sect at K, shew that each of the triangles A KB, DKC is isosceles.

11. If one angle of a triangle be equal to the sum of the other
two, the greatest side is double of the distance of its middle point
from the opposite angle.

THEOREMS AND EXAMPLES ON BOOK I.

1 2. Two right-angled triangles lohich have their hypotenuses equal,
and one side of one equxd to one side of the other, are equal in all
respects.

Let ABC, DEF be two A" right-angled at B and E, having AC
equal to DF, and AB equal to DE.

T/ien shall the A ABC he equal to the A DEF in all respects.

For apply the A ABC to the A DEF, so that AB may coincide
with the equal line DE, and C may fall on the side of DE remote
from F. Let C be the point on which C falls.

Then DEC represents the A ABC in its new position.

Now each of the L* DEF, DEC is a rt. L\ Hyp.

:. EF and EC are in one st. line. i. 14.

Then in the A CDF, because DF = DC {i.e. AC), Hyp.
:, the L DFC = the L DCF. i. 5.

Hence in the two A« DEF, DEC,
f the L DEF = the L DEC, being rt. Z." ;
Because \ and the Z. DFE =the Z. DCE ; Proved.

\ also the side DE is common to both ;
.*. the A» DEF, DEC are equal in all respects ; i. 26.
that is, the A» DEF, ABC are equal in all respects. q.e.d.

Alternative Proof. Since the L ABC is a rt. angle ;

.*. the sq. on AC = the sqq. on AB, BC. i. 47.

Similarly, the sq. on DF=the sqq. on DE, EF ; i. 47.

But the sq. on AC = the sq. on DF, since AC=DF ;
.*. the sqq. on AB, BC = the sqq. on DE, EF.

And of these, the sq. on AB = the sq. on DE, since AB = DE ;

,'. the sq. on BC = the sq. on EF ; Ax. 3.

:. BC = EF.

Hence the three sides of the A ABC are respectively equal to the
three sides of the A DEF ;

.-. the A ABC = the A DEF in all respects. i. 8.

100

eucud's elements.

13. If two triangles have two sides of the one eqttal to two sides
of the other, each to each, and hat>e likewise the angles opposite to one
pair of equal sides eqiial, then the angles opposite to the other pair of
equal sides shall be either equal or supplementary , and in the former
case the triangles shall he equal in all respects.

Let ABC, DEF be two triangles, in which
the side AB = the side DE,
the side AC = the side DF,
and the Z-ABC = the Z-DEF.

Then shaM the L* ACB, DFE 66 either equal (a« in Figs. 1 and 2)
or supplementary (a« in Figs. 1 and 3) ; ami in the former case the
triangles shall he eqvM in all respects.

If the L BAC = the L EDF. [Figs. 1 and 2.]

then the Z. ACB = the Z.DFE, and the triangles are equal in all

respects. i. 4.

But if the L BAC be not equal to the Z. EDF, [Figs. 1 and 3.]
let the Z- EDF be greater than the Z-BAC.

At D in ED make the L EDF' equal to the L BAC.
Then the A" BAC, EDF' are equal in all respects.

.-. AC=DF';
but AC=DF;
.-. DF = DF',
.-. the angle DFF'=the L DF'F.

But the L? DFT, DF'E are supplementary,
.'. the Z-" DFF', DF'E are supplementary :
that is, the Z-« DFE, ACB are supplementary.

I. 26.

Hyp,

I. 5.
I. 13.

Q.E.D.

Corollaries. Three cases of this theorem deserve special
attention.

It has been proved that if the angles ACB, DFE are not supple-
mentary they are eqtuxl :

THEOREMS AND EXAMPLES ON BOOK I. 101

Hence, in addition to the hypothesis of this theorem,

(i) If the angles ACB, DFE opposite to the two equal sides
AB, DE are both acute or both obtuse they cannot
be supplementary, and are therefore equal ; or if one
of them is a right angle, the other must also be a
right angle (whether considered as supplementary or
equal to it) :

in either case the triangles are equal in all respects.

(ii) If the two given angles are right angles or obtuse angles,
it follows that the angles ACB, DFE must be both
acute, and therefore equal, by (i) :

so that the triangles are equal in all respects.

(iii) If in each triangle the side opposite the given angle is
not less than the other given side ; that is, if AC and
DP are not less than AB and DE respectively) then
the angles ACB, DFE cannot be greater than the
angles ABC, DEF respectively ;

therefore the angles ACB, DFE are both acute ;

hence, as above, they are equal ;

and the triangles ABC, DEF are equal in all respects.

IL ON INEQUALITIES.
See Propositions 16, 17, 18, 19, 20, 21, 24, 25.

1. In a triangle ABC, if AC is not greater than AB, shew that
any straight line drawn through the vertex A, and terminated by the
base BC, is less than AB.

2. ABC is a tHangle^ and the vertical angle BAC is bisected by a
straight line which meets the base BC in X ; shew that BA is greater
than BX, and CA greater than CX. Hence obtain a proof of i. 20.

3. The perpendicular is the shortest straight line that can be
drawn from a given point to a given straight line ; and of others, that
which is nearer to the perpendicular is less than the more remote ; and
two, and only two eqval straight lines can be drawn from the given
point to the given straight line, one on each side of the perpendicular.

4. The sum of the distances of any point from the three angular
points of a triangle is greater than half its perimeter.

5. The sum of the distances of any point within a triangle from
its angular points is less than the perimeter of the triangle.

102 b:uclid*s elements.

6. The perimeter of a quadrilateral is greater than the sum of
its diagonals.

7. The sum of the diagonals of a quadrilateral is less than the
sum of the four straight lines drawn from the angular points to any
given point. Prove this, and point out the exceptional case.

8. In a triangle any two sides are together greater than tioice the
median which bisects the remaining side. [See Def. p. 79.]

[Produce the median, and complete the construction after the
manner of i. 16.]

9. In any triangle the sum of the medians is less than the peri-
meter,

10. In a triangle an angle is acute, obtuse, or a right angle,
according as the median drawn from it is greater than, less than, or
equal to half the opposite side. [See Ex. 4, p. 65.]

11. The diagonals of a rhombus are unequal.

12. If the vertical angle of a triangle is contained by unequal
sidesy and if from the vertex the median and the bisector of the angle
are drawnf then the median lies within the angle contained by the
bisector and the longer side.

Let ABC be a A, in which AB is greater
than AC ; let AX be the median drawn from
A, and AP the bisector of the vertical
L BAC.

Then shall AX lie between AP and AB.

Produce AX to K, making XK equal to
AX. Join KC.

Then the A» BXA, CXK may be shewn
to be equal in all respects ; i. 4.

hence BA = CK, and the L BAX = the L CKX.

But since BA is greater than AC, Hyp.
:. CK is greater than AC ;

/. the L CAK is greater than the L CKA : i. 18.

that is, the L CAX is greater than the L BAX :
.*. the L CAX must be more than half the vert. L BAC ;

hence AX lies within the angle BAP. q.e.d.

13. If the vertical angle of a triangle is contained by two unequal
sides f and if from the vertex there are drawn the bisector of the vertical
angle, the median, and the peipendicvlar to the base, the first of these
lines is intermediate in position and mugnitude to the other two.

THEOREMS AND EXAMPLES ON BOOK I. 103

III. ON PARALLELS.
See Propositions 27 — 3L

1. If a straight line meets two parallel straight lines, and the
two interior angles on the same side are bisected ; shew that the
bisectors meet at right angles, [i. 29, i. 32.]

2. The straight lines drawn from any point in the bisector of
an angle parallel to the arms of the angle, and terminated by them,
are equal ; and the resulting figure is a rhombus.

3. AB and CD are two straight lines intersecting at D, and the
adjacent ancles so formed are bisected : if through any point X in
DC a strai^t line YXZ be drawn parallel to AB and meeting the
bisectors in Y and Z, shew that XY is equal to XZ.

4. If two straight lines are parallel to two other straight lines,
each to each; and if the acute angles contained by each pair are
bisected ; shew that the bisecting lines are parallel.

5. The middle point of any straight line which meets two parallel
straight lines, and is terminated by them, is equidistant from the
parallels.

6. A straight line drawn between two parallels and terminated
by them, is bisected ; shew that any other straight line passing
through the middle point and terminated by the parallels, is also
bisected at that point.

7. If through a point equidistant from two parallel straight
lines, two straight lines are drawn cutting the parallels, the portions
of the latter thus intercepted are equal.

PROBLEMS.

8. AB and CD are two given straight lines, and X is a given
point in AB : find a point Y in AB such that YX may be equal to the
perpendicular distance of Y from CD.

9. ABC is an isosceles triangle : required to draw a straight
line DE parallel to the base BC, and meeting the equal sides in D
and E, so that BD, DE, EC may be all equal.

10. ABC is any triangle : required to draw a straight line DE
parallel to the base BC, and meeting the other sides in D and E, so
that DE may be equal to the sum of BD and CE.

11. ABC is any triangle : required to draw a straight line parallel
to the base BC, and meeting the other sides in D and E, so that DE
may be equal to the difference of BD and CE.

i04 fitJCLID^S ELEMENTS.

IV. ON PARALLELOGRAMS.

See Propositions 33, 34, and the deductions from these Props,
given on page 70.

1. The straight line drawn through the middle point of a side of a
triangle parallel to the base, bisects the remaining side.

Let ABC be a A, and Z the middle point of the side AB.
Through Z, ZY is drawn par^ to BC.

Then shall Y be the middle point of AC.

Through Z draw ZX par» to AC. i. 3L

Then in the A- AZY, ZBX,
because ZY and BC are par^
.-. the L AZY = the L ZBX ; i. 29.
and because ZX and AC are par^,
.-. the L ZAY=the L BZX ; i. 29.
alsoAZ = ZB: ffyp-

/. AY = ZX.

But ZXCY is a par™ by construction ;

.-. ZX = YC. 1.34.

Hence AY = YC;
that is, AC is bisected at Y. q.e.d.

2. The straight line which joins the middle points of two sides of a
triangle, is parallel to the third side.

Let ABC be a A, and Z, Y the middle
points of the sides AB, AC.

Then shall ZY be par^ to BC.

Produce ZY to V, making YV equal to

Join CV.
Then in the A- AYZ. CYV,

r AY = CY, Jlyp.

Because -j and YZ = YV, Constr.

land the L AYZ = the vert. opp. L CYV ; i. 15.

.-. AZ = CV, I. 4.

and the L ZAY = the L VCY ;

hence CV is par^ to AZ. i. 27.

But CV is equal to AZ, that is, to BZ : ^t/P-

.'. CV is equal and par^ to BZ :
.'. ZV is equal and par^ to BC : i. 33.

that is, Zy is par* to BC. Q.E.D.

[A second proof of this proposition may be derived from I. 38, 39.]

THEOREMS AND EXAMPLES ON BOOK I.

105

3. The straight line which joins the middle points of two sides of a
triangle is eqtial to half the third side.

4. Shew that the three straight lines which join the middle points
of the sides of a triangle, divide it into four triangles which are identi-
cally eqtuil.

5. Any straight line drawn from the vertex of a triangle to the
base is bisected by the straight line which joins the middle points of the
other sides of the triangle,

6. Given the three middle points of the sides of a triangle, con-
struct the triangle.

7. AB, AC are two given straight lines, and P is a given point
between them ; required to draw through P a straight line termi-
nated by AB, AC, and bisected by P.

8. ABCD is a parallelogram, and X, Y are the middle points of
the opposite sides AD, BC : shew that BX and DY trisect the dia-
gonal AC.

9. If the middle points of adjacent sides of any quadrilateral are
joined^ the figure thus formed is a parallelogram,

10. Shew that the straight lines which join the middle points of
opposite sides of a quadrilateral, bisect one another.

11. The straight line which joins the middle points of the oblique
sides of a trapezium, is parallel to the two parallel sides, and passes
through the middle points of the diagonals.

12. The straight line which joins the middle points of the oblique
sides of a trapezium is equal to half the sum of the parallel sides; and
the portion intercepted between the diagonals is eqtial to half the
difference of the parallel sides.

DEFINITION.

If from the extremities of one straight line perpendiculars are
drawn to another, the portion of the latter intercepted between
the perpendiculars is said to be the Orthogonal Projection of the first
line upon the second.

Thus in the adjoining figures, if from the extremities of the
straight line AB the perpendiculars AX, BY are drawn to PQ, then
XY is the orthogonal projection of AB on PQ.

106

Euclid's elements.

13. A given straight line AB is bisected at C ; shew that the pro-
jections o/AC, CB on any other straight line are equal.

X ^

p [^

Let XZ, ZY be the projections of AC, CB on any straight line PQ.
Then XZ and ZY shall he equal.

Through A draw a straight line parallel to PQ, meeting CZ, BY
or these lines produced in H, K. i. 31.

Now AX, CZ, BY are parallel, for they are perp. to PQ ; i. 28.
.*. the figures XH, HY are par°*»;

.-. AH = XZ, and HK = ZY. i. 34.

But through C, the middle point of AB, a side of the A ABK,
CH has been drawn parallel t-o the side BK ;

.-. CH bisects AK : Ex. 1, p. 104.

that is, AH = HK;

.*. XZ = ZY. Q.E.D.

14. If three parallel straight lines make equal intercepts on a
fourth straight line which meets thenij they will also make equal inter-
cepts on any other straight line which meets them.

15. Eqiud and parallel straight lines have eqival projections on any
other straight line.

] 6. AB is a given straight line bisected at O ; and AX, BY are
perpendiculars drawn from A and B on any other straight line : shew
that OX is equal to OY.

17. AB is a given straight line bisected at O : and AX, BY and OZ
are perpendiculars drawn to any straight litie PQ, which does not pass
between A and B : shew that OZ is eqiwtl to half the sum of AX, BY.

[OZ is said to be the Aritlixnetic Mean between AX and BY.]

18. AB is a given straight line bisected at O ; and through A, B
and O parallel straight lines are drawn to meet a given straight line
PQ in A, Y, Z : shew that OZ is equal to half the sum, or half the
difference of AX and BY, according as A and B lie on the same side
or on opposite sides of PQ.

THEOREMS AND EXAMlPLES ON BOOK I. 107

19. To divide a given finite straight line into any number of equal
parts.

[For example : required to divide the straight
line AB into five equal parts.

From A draw AC, a straight line of un-
limited length, making any angle with AB.

In AC take any point P ; and by marking off
successive parts PQ, QR, RS, ST each equal
to AP, make AT to contain kPfive times.

Join BT ; and through P, Q, R, S draw
parallels to BT.

It may be shewn by Ex. 14, p. 106, that
these parallels divide AB into five equal parts.]

20. If through an angle of a parallelogram any straight line
is dravmj the perpendicular drawn to it from the opposite angle is
equal to the sum or difference of the perpendiculars drawn to it from the
two remaining angles, according as the given straight line falls without
the parallelogram y or intersects it,

[Through the opposite angle draw a straight line parallel to the
given straight line, so as to meet the perpendicular from one of the
remaining angles, produced if necessary ; then apply i. 34, i. 26. Or
proceed as in the following example.]

21. From the angular points of a parallelogram perpendiculars
are drawn to any straight line which is without the parallelogram :
shew that the sum of the perpendiculars drawn from one pair of
opposite angles is equal to the sum of those drawn from the other pair.

[Draw the diagonals, and from their point of intersection let fall
a perpendicular upon the given straight line. See Ex. 17, p. 106.]

22. The sum of the perpendiculars drawn from any point in the
base of an isosceles triangle to the equal sides is equal to the perpen-
dicular drawn from either extremity of the base to the opposite side.

[It follows that the sum of the distances of any point in the base
of an isosceles triangle from the equal sides is constant, that is,
the same whatever point in the base is taken.]

23. In the base produced of an isosceles triangle any point is
taken : shew that the difference of its perpendicular distances from
the equal sides is constant.

24. The sum of the perpendiculars drawn from any point within
an equilateral triangle to the three sides is equal to the perpendicular
drawn from any one of the angular points to the opposite side, and
is therefore constant.

108 Euclid's elements.

PROBLEMS.

25. Draw a straight line through a given point, so that the part
of it intercepted between two given parallel straight lines may be of
given length. When does this problem admit of two solutions, when
of only one, and when is it impossible ?

26. Draw a straight line parallel to a given straight line, so
that the part intercepted between two other given straight lines
may be of given length.

27. Draw a straight line equally inclined to two given straight
lines that meet, so that the part intercepted between them may be
of given length.

28. AB, AC are two given straight lines, and P is a given point
wit/iout the angle contained by them. It is required to draw through
P a straight line to meet the given lines, so that the part intercepted
between them may be equal to the part between P and the nearer
line.

V. MISCELLANEOUS THEOREMS AND EXAMPLES.

Chiefly on i. 32.

1. A is the vertex of an isosceles triangle ABC, and BA is prodiiced
to D, 80 that AD is equLcU to BA ; if DC is draum, shew that BCD is a
right angle.

2. The straight line joining the middle point of the hypotemise of
a right-angled triangle to the right angle is equal to half the hypotenuse,

3. From the extremities of the base of a triangle perpendiculars
are drawn to the opposite sides (produced if necessary) ; shew that
the straight lines which join the middle point of the base to the feet
of the perpendiculars are equal.

4. In a triangle ABC, AD is drawn perpendicular to BC ; and
X, Y, Z are the middle points of the sides BC, CA, AB respectively :
shew that each of the angles ZXY, ZDY w equal to the angle BAC.

5. In a right-angled triangle^ if a perpendicular is drawn from
the right angle to the hypotenuse, the tv)o triangles thus formed are
equiangidar to one another.

6. In a right-angled triangle two straight lines are drawn from
the right angle, one bisecting the hypotenuse, the other perpendicular
to it : shew that they contain an angle equal to the difference of the tujo
acute angles of the triangle, [See above, Ex. 2 and Ex. 6. J

THEOREMS AND EXAMPLES ON BOOK I. 109

7. In a triangle if a perpendicular is drawn from one extremity
of the base to the bisector of the vertical angle, (i) it will make with
either of the sides containing the vertical angle an angle equal to half
the sum of the angles at the ba^se; (ii) it will make with the base an
angle equal to half the difference of the angles at the base.

Let ABC be the given A, and AH the bi-
sector of the vertical L BAG.

Let CLK meet AH at right angles.

(i) Then shall each of the Z-« AKC, ACK
be equal to half the sum of the Z-« ABC,
ACS

In the A- AKL, ACL, B X H C

r the L KAL=the L CAL, Hyp.

Because also the L ALK = the L ALC, being rt, L* ;

and AL is common to both A» ;
.-. the L AKL = the L ACL. i. 26.

Again, the L AKC = the sum of the Z.« KBC, KCB ; i. 32.
.-. the L ACK = the sum of the Z-» KBC, KCB.
To each add the L ACK :
then twice the L ACK = the sum of the L* ABC, ACB ;

.-. the L ACK = half the sum of the Z.« ABC, ACB.

(ii) The L KCB shaU be equal to half the difference of the
Z.« ACB, ABC.

As before, the L ACK = the sum of the L* KBC, KCB.

To each of these add the L KCB :

then the L ACB = the Z. KBC together with twice the L KCB.

.-. twice the L KCB = the difference of the Z-« ACB, KBC ;

that is, the L KCB = half the difiference of the Z.» ACB, ABC.

^ Corollary. If Yk is the middle point of the base^ and XL is
joined, it may be shewn by Ex. 3, p. 105, that XL is half BK; that is,
thai XL is half the difference of the sides AB, AC.

8. In any triangle the angle contained by the bisector of the vertical
angle and the perpendicular from the vertex to the base is equal to half
the difference of the angles at the base. [See Ex. 3, p. 65.]

9. Li a triangle ABC the side AC is produced to D, and the
angles BAC, BCD are bisected by straight lines which meet at F ;
dhew that they contain an angle equal to half the angle at B.

10. If in a right-angled triangle one of the acute angles is double
of the other, shew that the hypotenuse is double of the shorter side.

110 Euclid's elements.

11. If in a diagonal of a parallelogram any two points equi-
distant from its extremities are joined to the opposite angles, the
figure thus formed will be also a parallelogram.

12. ABC is a given equilateral triangle, and in the sides BC, CA,
AB the points X, Y, Z are taken respectively, so that BX, CY and AZ
are all equal. AX, BY, CZ are now drawn, intersecting in P, Q, R :
shew that the triangle PQR is equilateral.

13. If in the sides AB, BC, CD, DA of a parallelogram ABCD
four points P, Q, R, S are taken in order, one in each side, so that
AP, BQ, CR, DS are all equal ; shew that the figure PQRS is a
parallelogram.

14. In the figure of i. 1, if the circles intersect at F, and if CA
and CB are produced to meet the circles in P and Q respectively ;
shew that the points P, F, Q are in the same straight line; and
shew also that the triangle CPQ is equilateral.

[Problems marked (*) admit in general of more than one solution.]

15. Through two given points draw two straight lines forming
with a straight line given in position, an equilateral triangle.

*16. From a given point it is required to draw to two parallel
straight lines two equal straight lines at right angles to one another.

*17. Three given straight lines meet at a point ; draw another
straight line so that the two portions of it intercepted between the
given lines may be equal to one another.

18. From a given point draw three straight lines of given lengths,
so that their extremities may be in the same straight line, and inter-
cept equal distances on that line. [See Fig. to i. 16.]

19. Use the properties of the equilateral triangle to trisect a
given finite straight line.

20. In a given triangle inscribe a rhombus, having one of its
angles coinciding with an angle of the triangle.

VI. ON THE CONCURRENCE OF STRAIGHT LINES IN A

TRIANGLE.

Definitions, (i) Three or more straight lines are said to be
concurrent when they meet in one point.

(ii) Three or more points are said to be colllnear when they lie
upon one straight line.

Obs. We here give some propositions relating to the concurrence
of certain groups of straight lines drawn in a triangle : the import-
ance of these theorems will be more fully appreciated when the
student is familiar with Books iii. and iv.

THEOREMS AND EXAMPLES ON BOOK I. Ill

1. The perpendictUars drawn to the sides of a triangle from their
middle points are concurrent.

Let ABC be a A, and X, Y, Z the
middle points of its sides.

Then shall the perp* drawn to the sides
from X, Y, Z he concurrent.

From Z and Y draw perp" to AB, AC ;
these perp", since they cannot be parallel,
will meet at some point O. Ax. 12.

Join OX.

It is required to prove that OX is perp. to BC.

Join OA, OB, OC.
In the A» OYA, OYC,
[ YA = YC, Hyp.

Because -[ and OY is common to both ;

(also the Z. OYA = the L OYC, being rt. L* ;

:. OA=OC. I. 4.

Similarly, from the A" OZA, OZB,

it may be proved that OA = OB.

Hence OA, OB, OC are all equal.

Again, in the A» OXB, OXC,
r BX = CX, Hyp,

Because -! and XO is common to both ;

[ also OB = OC : Proved.

.'. the L OXB = the L OXC ; i. 8.

but these are adjacent L* ;

.'. they are rt. L* ; Def. 10.

. that is, Ox is perp. to BC.

Hence the three perp» OX, OY, OZ meet at the point O.

Q. E. D.

2. The bisectors of the angles of a triangle are concurrent.

Let ABC be a A. Bisect the Z.« ABC,
BOA, by sti aight lines which must meet
at some point O. Ax. 12.

Join AO.
It is required to prove that AO bisects the

L BAC.
From O draw OP, OQ, OR perp. to the
sides of the A.

Then in the A« OBP, OBR,
r the L OBP=the L OBR,

Because -j and the L OPB = the L ORB, being rt. Z-»,
\ and OB is common ;

.-. OP=OR. I. 26

112

EUCLID'S ELEMENTS.

Similarly from the A« OOP, OCQ,
it may be shewn that OP = OQ,
.'. OP, OQ, OR are all equal.

Again in the A» ORA, OQA,

(the Z." ORA, OQA are rt. L\
and the hypotenuse OA is
common,
also OR = OQ ; Proved.
.-. the L RAO = the L QAO. Ex. 12, p. 99.

That is, AO is the bisector of the L BAG.

Hence the bisectors of the three L* meet at the point O.

Q.E.D.

3. T?ie bisectors of ttco exterior angles of a triangle and the
bisector of the third angle are concurrent.

Let ABC be a A, of which the sides AB,
AC are produced to any points D and E.

Then shall the bisectors of the L* DBC,
ECB, BAC be concvrrent.

Bisect the Z.» DBC, ECB by straight lines
which must meet at some point O. Ax. 12.

Join AO.

It is remdred to prove thai AO Insects the
angle BAC.

From O draw OP, OQ, OR perp. to the
sides of the A.

Then in the A» OBP, OBR,

r theZ-OBP=theZ-OBR,
Because-^ also the L OPB = the L ORB, being rt. L\
I and OB is common ;

.-. OP=OR. I. 26.

Similarly from the A» OCP, OCQ,

it may be shewn that OP = OQ :

.-. OP, OQ, OR are all equal.

Again in the A» ORA, OQA,
r the Z-" ORA, OQA are rt. L\
Because -I and the hypotenuse OA is common,

\ alsoOR = OQ; Proved.

:. the L RAO = the L QAO. Ex. 12, p. 99.

That is, AO is the bisector of the L BAC.

.*. the bisectors of the two exterior Z-» DBC, ECB,

and of the interior L BAC meet at the point O.

Q.E.D.

THEOREMS AND EXAMPLES ON BOOK I. 113

4. The medians of a triangle are concurrent.

Let ABC be a A.
T?ien shcUl its three medians he concurrent.
Let BY and CZ be two of its medians, and
let them intersect at O.

Join AO,

and produce it to meet BC in X.

It is required to shew that AX is the remaining

median of the A.

Through C draw CK parallel to BY :

produce AX to meet CK at K.

Join BK, ^J^

In the A AKC,
because Y is the middle point of AC, and YO is parallel to CK,

.'. O is the middle point of AK. Ex. 1, p. 104.

Again in the A ABK,
since Z and O are the middle points of AB, AK,

/. ZO is parallel to BK, Ex. 2, p. 104.

that is, OC is parallel to BK :

.'. the figure BKCO is a par".

But the diagonals of a par*" bisect one another, Ex. 5, p. 70.

.*. X is the middle point of BC,

That is, AX is a median of the A.

Hence the three medians meet at the point O. q.e.d.

Corollary. The three medians of a triangle cut one another at a
point of trisection, the greater segment in each being towards the
angular point.

For in the above figure it has been proved that

AO = OK,

also that OX is half of OK ;

.-. OX is half of OA :

that is, OX is one third of AX.

Similarly OY is one third of BY,

and OZ is one third of CZ. Q.E.D.

By means of this Corollary it may be shewn that in any triangle
the shorter median bisects the greater side.

[The point of intersection of the three medians of a triangle is
called the centroid. It is shewn in Mechanics that a thin triangular
plate will balance in any position about this point : therefore the
centroid of a triangle is also its centre of gravity.]
H.S.S. H

114

EUCLID'S ELEMENTS.

5. The perpeiidictUars drawn from the vertices of a triangle to the
opposite sides are concurrent.

Let ABC be a A, and AD, BE, CF the three perp" drawn from
the vertices to the opposite sides.

Then shall the perp' AD, BE, CF he concurrent.

Through A, B, and C draw straight lines MN, NL, LM parallel
to the opposite sides of the A.

Then the figure BAMC is a par™. Def. 36.

.-. AB = MC. I. 34.

Also the figure BACL is a par™.
.-. AB = LC,
.-. LC=CM:
that is, C is the middle point of LM.
So also A and B are the middle points of M N and N L.
Hence AD, BE, CF are the perp" to the sides of the A LMN from
their middle points. Ex. 3, p. 60.

But these perp" meet in a point : Ex. 1, p. 111.
that is, the perp" drawn from the vertices of the A ABC to the
opposite sides meet in a point. Q.E.D.

[For another proof see Theorems and Examples on Book iii.]

DEFINITIONS.

(i) The intersection of the perpendiculars drawn from the
vertices of a triangle bo the opposite sides is called its ortbocentre.

(ii) The triangle formed by joining the feet of the perpen-
4iculars is called the pedal trian^lQ.

THEOREMS AND EXAMPLES ON BOOK I.

115

VIL ON THE CONSTRUCTION OF TRIANGLES WITH GIVEN

PARTS.

Ohs. No .general rules can be laid down for the solution of
problems in this section ; but in a few typical cases we give con-
structions, which the student will find little difficulty in adapting
to other questions of the same class.

1. Construct a right-angled triangle^ having given the hypotemise
and the sum of the remaining sides,

[It is required to construct a rt.-
angled A, having its hypotenuse equal
to the given straight line K, and the sum
of its remaining sides equal to AB.

From A draw AE making with BA
an L equal to half a rt. Z-. From
centre B, with radius equal to K, de-
scribe a circle cutting Ac in the points

c, c

From C and C draw perp» CD, CD' to AB ; and join CB, C'B.
Then either of the A" CDB, C'D'B will satisfy the given conditions.

Note. If the given hypotenuse K be greater than the perpen-
dicular drawn from B to AE, there will be two solutions. If the line
K be equal to this perpendicular, there will be one solution ; but if
less, the problem is impossible.]

2. Construct a right-angled triangle, having given the hypotenuse
and the difference of the remaining sides.

3. Construct an isosceles right-angled triangle, having given the
sum of the hypotenuse and one side.

4. Construct a triangle y having given the perimeter and the angles
at the base.

[Let AB be the perimeter of the required A, and X and Y the Z." at
the base.

From A draw AP, making the L BAP equal to half the L X.
From B draw BP, making the Z. ABP equal to half the L Y.
From P draw PQ, making the L APQ equal to the L BAP.
From P draw PR, making the L BPR equal to the L ABP.
Then shall PQR be the required A.]

116 Euclid's elements.

5. Construct a right-angled triangle, having given the perimeter
and one acute angle.

6. Construct an isosceles triangle of given altitude, so that its
base may be in a siven straight line, and its two equal sides may
pass through two fixed points. [See Ex. 7* p. 55.]

7. Construct an equilateral triangle, having given the length of
the perpendicular drawn from one of the vertices to the opposite side.

8. Construct an isosceles triangle, having given the base, and
the difference of one of the remaining sides and the perpendicular
drawn from the vertex to the base. [See Ex. 1, p. 96.]

9. Construct a triangle, having given the base, one of the angles
at the base, and the sum of the remaining sides.

10. Construct a triangle, having given the base, one of the angles
at the base, and the difference of the remaining sides. [Two cases
arise, according as the given angle is adjacent to the greater side or
the less.]

11. Construct a triangle^ having given the 6a»e, the difference of
the angles at the hose, and the difference of the remaining sides.

[Let AB be the given base, X the difference of the Z.* at the base,
and K the difference of the remaining sides.

Draw BE, making the L ABE equal to half the L X.
From centre A, with radius equal to K, describe a circle cutting
BE in D and D'. Let D be the point of intersection nearer to B.

Join AD and produce it to 0.
Draw BO, making the L DBG equal to the L BDC.

Then shall CAB be the A required. Ex. 7, p. 109.

Note. This problem is possible only when the given difference
K is greater than the perpendicular drawn from A to BE.]

12. Construct a triangle, having given the base, the difference
of the angles at the base, and the sum of the remaining sides.

13. Construct a triangle, having given the perpendicular from
the vertex on the base, and the difference between each side and the
adjacent segment of the base.

THEOREMS AND EXAMPLES ON BOOK I. 117

14. Construct a triangle, having given two sides and the median
which bisects the remaining side. [See Ex. 18, p. 110.]

15. Construct a triangle, having given one side, and the medians
which bisect the two remaining sides.

[See Fig. to Ex. 4, p. 113.

Let BO be the given side. Take two-thirds of each of the given
medians; hence construct the triangle BOG. The rest of the con-
struction follows easily.]

16. Construct a triangle y having given its three medians,

[See Fig. to Ex. 4, p. 113.

Take two- thirds of each of the given medians, and construct the
triangle OKO. The rest of the construction follows easily.]

VIII. ON AREAS.
See Propositions 35 — 48.

Ohs. It must be understood that throughout this section the
word eqtud as applied to rectilineal figures will be used as denoting
equality of area unless otherwise stated,

1. Shew that a parallelogram is bisected by any straight line
which parses through the middle point of one of its diagonals.

[I. 29, 26.]

2. Bisect a parallelogram by a straight line drawn through a
given point.

3. Bisect a parallelogram by a straight line drawn perpendicular
to one of its sides.

4. Bisect a parallelogram by a straight line drawn parallel to a
given straight line.

5. ABOD is a trapezium in which the side AB is parallel to DC.
Shew that its area is equal to the area of a parallelogram formed by
drawing through X, the middle point of BC, a straight line parallel to
AD, meeting DC, or DC produced. [i. 29, 26.]

6. A trapezium is equal to a parallelogram whose base is half the
sum of the parallel sides of the given figure, and whose altitude is
equal to the perpendicular distance between them.

7. ABCD is a trapezium in which the side AB is parallel to
DC ; shew that it is double of the triangle formed by joining the
extremities of AD to X, the middle point of BC.

8. Shew that a trapezium is bisected by the straight line which
joins the middle points of its parallel sides. [i. 38«\

118 Euclid's elements.

Obs. In the following group of Exercises the proofs depend
chiefly on Propositions 37 and 38, and the two converse theorems.

9. If two straight lines AB, CD intersect at X, and if the
straight lines AC and BD, which join their extremities are parallel,
shew that the triangle AXD is equal to the triangle BXC.

10. If two straight lines AB, CD intersect at X, so that the
triangle AXD is equal to the triangle XCB, then AC and BD are
parallel.

IL ABCD is a parallelogram, and X any point in the diagonal
AC produced ; shew that the triangles XBC, a DC are equal. [See
Ex. 13, p. 70.]

12. ABC is a triangle, and R, Q the middle points of the sides
AB, AC ; shew that if BQ and CR intersect in X, the triangle BXC
is equal to the quadrilateral AQXR. [See Ex. 5, p. 79.]

13. If the middle points of the sides of a quadrilateral be joined
in order, the parallelogram so formed [see Ex. 9, p. 105] is equal to
half the given figure.

14. Two triangles of equal area stand on the same base but on
opposite sides of it : shew that the straight line joining their vertices
is bisected by the base, or by the base produced.

15. The straight line which joins the middle points of the dia-
gonals of a trapezium is parallel to each of the two parallel sides.

16. (i) A triangle is equal to the sum or difference of two triangles
on the same base {or on equal basses), if the altitude of the first is equal
to the sum or difference of the altitudes of the others.

(ii) A triangle is equal to the sum or difference of two triangles of
the same altitudey if the base of the first is equal to the sum or difference
of the ba^es of the others.

Similar statements hold good of parallelograms.

17. ABCD is a parallelogram, and O is any point outside it;
shew that the sum or difference of the triangles OAB, OCD is equal
to half the parallelogram. Distinguish between the two cases.

Obs. On the following proposition depends an important theorem
in Mechanics : we cive a proof of the first case, leaving the second
case to be deduced oy a similar method.

THEOREMS AND EXAMPLES ON BOOK 1. 119

18. (i) ABCD 18 a paralldo<jram^ and O is any point without
the angle BAD and its opposite vertical angle ; shew that the triangle
OAC w equal to the sum q/* the triangles OAD, OAB.

(ii) If O is within the angle BAD or its opposite vertical ancle,
the triangle OAC is equal to the difference of the triangles OAD, OAB.

Case I. If O is without the L DAB
and its opp. vert. L, then OA is with-
out the par"* ABCD : therefore the perp.
drawn from C to OA is eoual to the sum
of the perp" drawn from B and D to OA.
[See Ex. 20, p. 107.]

Now the A» OAC, OAD, OAB are
upon the same base OA ;
and the altitude of the A OAC with
respect to this base has been shewn to
be equal to the sum of the altitudes of
the A- OAD, OAB.
Therefore the A OAC is equal to the sum of the A> OAD, OAB.
[See Ex. 16, p. 118.] q.e.d.

19. ABCD is a parallelogram, and through O, any point within
it, straight lines are drawn parallel to the sides of the parallelogram ;
shew that the difference of the parallelograms DO, BO is douole of
the triangle AOC. [See preceding theorem (ii).]

20. The area of a quadrilateral is equal to the area of a triangle
having two of its sides equal to the diagonals of the given figure, and
the included angle equal to either of the angles between the dia-
gonals.

21. ABC is a triangle^ and D is any point in AB ; it is required to
draw through D a straight line DE. to meet BC produced m E, w that
the triangle DBE may he equal to the triangle ABC.

[Join DC. Through A draw AE parallel to DC. i. 31.

Join DE.

The A EBD shall be equal to the A ABC] i. 37.

120 EtTCLlD*S ELEMENTS.

22. On a base of given length describe a triangle equal to ft
given triangle and having an angle equal to an angle of the given
triangle.

23. Construct a triangle equal in area to a given triangle, and
having a given altitude.

24. On a base of given length construct a triangle equal to a
given triangle, and having its vertex on a given straight line.

25. On a base of given length describe (i) an isosceles triangle ;
(ii) a right-angled triangle, equal to a given triangle.

26. Construct a triangle equal to the sum or difference of two
given triangles. [See Ex. 16, p. 118.]

27. ABC is a given triangle, and X a given point : describe a
triangle equal to ABC, having its vertex at X, and its base in the
same straight line as BC.

28. ABCD is a quadrilateral. On the hose AB construct a triangle
equal in area to AdCD, and having the angle at A common with the
qiuidrilateral.

[Join BD. Through C draw CX parallel to BD, meeting AD
produced in X ; join da.]

29. Construct a rectilineal Jlgxvre equal to a given rectilineal
fgure^ and having fewer sides by one than the given figure.

Hence shew how to construct a triangle equal to a given rectilineal
figure.

30. ABCD is a quadrilateral : it is required to construct a triangle
equal in area to ABCD, having its vertex at a given point X in DC,
and its base in the same straight line as AB.

31. Construct a rhombus equal to a given parallelogram.

32. Construct a parallelogram which shall have the same area
and perimeter as a given triangle.

33. Bisect a triangle by a straight line drawn through one of its
angular points.

34. Trisect a triangle by straight lines drawn through one of its
angular points. [See Ex. 19, p. 110, and i. 38.]

35. Divide a triangle into any number of equal parts by straight
lines drawn through one of its angular points.

[See Ex. 19, p. 107, and i. 38.]

THEOREMS AND EXAMPLES ON BOOK I. 121

36. Bisect a triangle by a straight line dravm thrmigh a given
point in one of its sides.

[Let ABC be the given A, and P the
given point in the side AB.

Bisect AB at Z ; and join CZ, CP.
Through Z draw ZQ parallel to CP.
Join PQ.
Then shall PQ bisect the A. '

See Ex. 21, p. 119.] 8

37. Trisect a triangle by straight lines draion fr<ym a given point
in one of its sides.

[Let ABC be the given A, and X the
given point in the side BC.

Trisect BC at the points P, Q. Ex. 19, p. 107.
Join AX, and through P and Q draw PH
and QK parallel to AX.

Join XH, XK.
These straight lines shall trisect the A ; as
may be shewn by joining AP, AQ. ^f p — y A

See Ex. 21, p. 119.] ^ »- a w

38. Cut off from a given triangle a fourth, fifth, sixth, or any
part required by a straight line drawn from a given point in one of
its sides. [See Ex. 19, p. 107, and Ex. 21, p. 119.]

39. Bisect a quadrUatercd by a straight line drawn through an
angular point.

[Two constructions may be given for this problem : the first will
be suggested by Exercises 28 and 33, p. 120.

The second method proceeds thus.

Let ABCD be the given quadri-
lateral, and A the given angular point.

Join AC, BD, and bisect BD in X.
Through X draw PXQ parallel to AC,

meeting BC in P ; join A P.
Then shall AP bisect the quadrilateral.

Join AX, CX, and use i. 37, 38.]

40. Cut off from a given quadrilateral a third, a fourth, a fifth,
or any part required, by a straight line drawn through a given
angular point. [See Exercises 28 and 35, p. 120.]

122 EUCLID'S ELEMENTS.

Ohs, The following Theorems depend on i. 47.

41. In the figure of i. 47, shew that

(i) the sum of the squares on AB and AE is equal to the sum
of the squares on AC and AD.

(ii) the square on EK is equal to the square on AB with four
times the square on AC.

(iii) the sum of the squares on EK and FD is equal to five
times the square on BC.

42. If a straight line is divided into any two parts, the square
on the straight line is greater than the sum of the squares on the
two parts.

43. If the square on one side of a triangle is less than the
squares on the remaining sides, the angle contained by these sides is
acute ; if greater, obtuse.

44. ABC is a triangle, right-angled at A ; the sides AB, AC are
intersected by a straight line PQ, and BQ, PC are joined : shew
that the sum of the squares on BQ, PC is equal to the sum of the
squares on BO, PQ.

45. In a right-angled triangle four times the sum of the squares
on the medians which bisect the sides containing the right angle is
equal to five times the square on the hypotenuse.

46. Describe a square whose area shall be three times that of a
given square.

47. Divide a straight line into two parts such that the sum of
their squares shall be equal to a given square.

IX. ON LOCI.

In many geometrical problems we are required to find the position
of a point which satisfies given conditions ; and all such problems
hitherto considered have been found to admit of a limited number of
solutions. This, however, will not be the case if only one condition
is given. For example :

(i) Required a point which shall be at a given distance from a
given point.

This problem is evidently indeterminate^ that is to say, it admits
of an indefinite number of solutions ; for the condition stated is
satisfied by any point on the circumference of the circle described
from the given point as centre, with a radius equal to the- given
distance. Moreover this condition is satisfied by no other point
within or without the circle.

THEOREMS AND EXAMPLES ON BOOK I.

123

(ii) Required a point wldch shall he at a given distance from a
given straight line.

Here again there are an infinite number of such points, and they
lie on two parallel straight lines drawn on either side of the given
straight line at the given distance from it : further, no point that is
not OD one or other of these parallels satisfies the given condition.

Hence we see that one condition is not sufficient to determine the
position of a point absolutely, but it may have the effect of restrict-
ing it to some definite line or lines, straight or curved. This leads
us to the following definition.

Definition. The Locus of a point satisfying an assigned con-
dition consists of the line, lines, or part of a line, to which the point
is thereby restricted ; provided that the condition is satisfied by
every point on such line or lines, and by no other.

A locus is sometimes defined as the path traced out by a point
which moves in accordance with an assigned law.

Thus the locus of a point, which is always at a given distance
from a given point, is a circle of which the given point is the centre :
and the locus of a point, which is always at a given distance from a
given straight line, is a pair of parallel straight lines.

We now see that in order to infer that a certain line, or system
of lines, is the locus of a point under a given condition , it is necessary
to prove

(i) that any point which fulfils the given condition is on the
supposed locus ;

(ii) that every point on the supposed locus satisfies the given
condition.

1. Find the locus of a point which is always equidistant from two
given points.

Let A, B be the two given points,
(o) Let P be any point equidistant from A
and B, sothat AP = BP.

Bisect AB at X, and join PX.
Then in the A" AXP, BXP,
C AX = BX,

Because \ and PX is common to both,
\ alsoAP = BP,

.-. the L PXA = the Z. PXB ;
and they are adjacent L? ;

.-. PX is perp. to AB. Def. 10.
.'. any point which is equidistant from A and B
is on the straight line which bisects AB at right angles.

Constr.

Hyp,
I. 8.

124 EUCLID'S ELEMENTS.

()8) Also every point in this line is equidistant from A and B.

For let Q be any point in this line.

Join AQ, BQ.

Then in the A- AXQ, BXQ,

r AX = BX,

Because < and XQ is common to both ;

(also the L AXQ = the L BXQ, being rt. Z-»;

.-. AQ=BQ. 1.4.

That is, Q is equidistant from A and B.

Hence we conclude that the locus of the point equidistant from
two given points A, B is the straight line which bisects AB at right
angles.

2. To Jlnd the locits of the middle point of a straight line drawn
from a given point to meet a given straight line of unlimited length.

Let A be the given point, and BC the given straight line of un-
limited length.

(a) Let AX be any straight line drawn through A to meet BC,
and let P be its middle point.

Draw AF perp. to BC, and bisect AF at E.
Join Er, and produce it indefinitely.

Since AFX is a A, and E, P the middle points of the two sides AF, AX,
/. EP is parallel to the remaining side FX. Ex. 2, p. 104.
.*. P is on the straight line which passes through the fixed point E,
and is parallel to BC.

(/3) Again, every point in E P, or E P produced, fulfils the required
condition.

For, in this straight line take any point Q.
Join AQ, and produce it to meet BC in Y.

Then FAY is a A, and through E, the middle point of the side AF,
EQ is drawn parallel to the side FY ;

.*. Q is the middle point of AY. Ex. 1, p. 104.

Hence the required locus is the straight line drawn parallel to BC,
and passing through E, the middle point of the perp. from A to BC.

THEOREMS AND EXAMPLES ON BOOK I. 125

3. Find the locos of a point equidistant from two given inter-
secting straight lines. [See Ex. 3, p. 65.]

4. Find the locus of a point at a given radial distance from the
circumference of a given circle.

5. Find the locus of a point which moves so that the sum of its
distances from two given intersecting straight lines of unlimited
length is constant.

6. Find the locus of a point when the differences of its distances
from two given intersecting straight lines of unlimited length is
constant.

7. A straight rod of given length slides between two straight
rulers placed at right angles to one another : find the locus of its
middle point. [See Ex. 2, p. 108.]

8. On a given base as hypotenuse right-angled triangles are
described : find the locus of their vertices. [See Ex. 2, p. 108.]

9. AB is a given straight line, and AX is the perpendicular drawn
from A to any straight line passing through B : find the locus of the
middle point of AX.

10. Find the locus of the vertex of a triangle, when the base and
area are given.

11. Find the locus of the intersection of the diagonals of a
parallelogram, of which the base and area are given.

12. Find the locus of the intersection of the medians of tri-
angles described on a given base and of given area.

X. ON THE INTERSECTION OF LOCI.

It appears from various problems which have already been con-
sidered, that we are often required to find a point, the position of
which is subject to two given conditions. The method of loci is
very useful in solving problems of this kind ; for corresponding to
each condition there will be a locus on which the required point
must lie. Hence all points which are common to these two loci,
that is, all the points of intersection of the loci, will satisfy both the
given conditions.

126 EUCLID'S ELEMENTS.

Example I. To construct a triangle, having given the ha>se, the
altitude^ and the length of the median which bisects the base.

Let AB be the given base, and P and Q the lengths of the altitude
and median respectively :

then the triangle is known if its vertex is known.

(i) Draw a straight line CD parallel to AB^ and at a distance
from it equal to P :

then the required vertex must lie on CD.

(ii) Again, from the middle point of AB as centre, with radius
equal to Q, describe a circle :

then the required vertex must lie on this circle.

Hence any points which are common to CD and the circle,
satisfy both the given conditions : that is to say, if CD intersect the
circle in E, F each of the points of intersection might be the vertex
of the required triangle. This supposes the length of the median
Q to be greater than the altitude.

Example 2. To find a point equidistant from three given points
A, B, C, which are not in the same straight line,

(i) The locus of points equidistant from A and B is the straight
line PQ, which bisects AB at right angles. Ex. 1, p. 123.

(ii) Similarly the locus of points equidistant from B and C is
the straight line RS which bisects BC at right angles.

Hence the point common to PQ and RS must satisfy both con-
ditions : that is to say, the point of intersection of PQ and RS will
be equidistant from A, B, and C.

Obs. These principles may also be used to prove the theorems
relating to concurrency already given on page 111.

Example. To prove that the bisectors of the angles of a triangle
are concurrent.

Let ABC be a triangle.
Bisect the Z." ABC, BCA by straight
lines BO, CO : these must meet at
some point O. Ax. 12.

Join OA.

Then shall OA bisect the L BAC.
Now BO is the locus of points equi-
distant from BC, BA ; Ex. 3, p. 65.
/. OP=OR.
Similarly CO is the locus of points equi-
distant from BC, CA.

/. OP = OQ; hence OR = OQ.

/. O is on the locus of points equidistant from AB and AC :

that is, OA is the bisector of the L BAC.

Hence the bisectors of the three Z." meet at the point O.

THEOREMS AND EXAMPLES ON BOOK I. 127

It may happen that the data of the problem are so related to one
another that the resulting loci do not intersect. In this case the
problem is impossible.

For example, if in Ex. 1, page 126, the length of the given
median is less than the given altitude, the straight line CD will not
be intersected by the circle, and no triangle can fulfil the conditions
of the problem. If the length of the median is equal to the given
altitude, one point is common to the two loci ; and consequently
only one solution of the problem exists : and we have seen that
there are two solutions, if the median is greater than the altitude.

In examples of this kind the student should make a point of
investigating the relations which must exist among the data, in
order that the problem may be possible ; and he must observe that
if under certain relations two solutions are possible, and uuder other
relations no solution exists, there will always be some intermediate
relation under which one and only one solution is possible.

EXAMPLES.

1. Find a point in a given straight line which is equidistant
from two given points.

•

2. Find a point which is at given distances from each of two
given straight lines. How many solutions are possible ?

3. On a given 6owe covMnict a triangle^ having given one angle at
the ha^e and the length of the opposite side. Examine the rdations
which must exist among the data in order that there may he two
solvtionsy one solution^ or that the problem may he impossible.

4. On the base of a given triangle construct a second triangle
equal in area to the first, and having its vertex in a given straight
line.

5. Construct an isosceles triangle equal in area to a given
triangle, and standing on the same base.

6. Find a point which is at a given distance from a given point,
as^ is equidistant from two given parallel straight lines.

When does this problem admit of two solutions, when of one
only, and when is it impossible ?

BOOK II.

Book II. deals with the areas of rectangles and squares.

A Rectangle has been defined (Book I., Def. 37) as a
parallelogram which has one of its angles a right angle.

It should be remembered that if a parallelogram has one right
angle, all its angles are right angles. [i. 46, CiyrJ]

Definitions.

1. A rectangle is said to be contained by any two of
its sides which form a right angle : for it is clear that both
the form and magnitude of a rectangle are fully determined
when the lengths of two such sides are given.

Thus the rectangle ACDB is said
to he contained by AB, AC ; or by CD,
DB : and if X and Y are two straight
lines equal respectively to AB and AC,
then the rectangle contained by X and Y
is equal to the rectangle contained by
AB, AC.

[See Ex. 12, p. 70.]

X-
Y-

After Proposition 3, we shall use the abbreviation
rect AB, AC to denote the rectangle contained by AB and
AC.

2. In any parallelogram the figure formed by either
of the parallelograms about a diagonal together with the
two complements is called a gnomon.

Thus the shaded portion of the annexed
diagram, consisting of the parallelogram
EH together with the complements AK,
KC is the gnomon AHF.

The other gnomon in the diagram is
that which is made up of the figures AK,
GF and FH, namely the gnomon AFH.

BOOK 11. INTRODUCrORy.

Before eulerfaig^ upon Book U. the student is reminded of the
following arithmetical rule :

RoLB. To ^nd Ihe area of a rectangle, mvitipiy the number oj
imtff in thi Isnctli hy the mtmber qfvnita in tht breadtti; the produci
tuill be the mimber 0/ square units in the area.

For example, if the two sideB AB, AD a B

of the rectangle ABCD are respectively
/our and three inches long, and if through
the points of division parallels are drawn
aa in the annexed fieure, it is seen that
the rectangle ia divided into three rows,
eooh containing /our square inohea, or into
four columns, each containing thrtt square
inches. -

Hence the whole rectangle containn 3x4, or 12, eqnare Incheo.

Similarly if AB and AD contain m and n unita of length
respectively, it follows that the rectande ABCD will contftin mxn
units of area : further, if AB and AD are equal, each containing
m unics of length, the rectangle becomes a square, and contains

From thie we conclude that the rectaTigle contained by two straight
lines in Geometry corresponds to the product of tvio numiers in
Arithmetic or Algebra ; and that the square described on a atroighZ
litie corresponds to the square of a num6w. Accordingly it will be
found in the course of Book II, that several theorems relating to
the areas of rectangles and squares are analogous b> well-known
algebraical formulEe.

In view of these principles the rectangle contaiued by two
straight lines AB, BC is sometimea Bxpressed in the form of a
mthduct, as AB . BC, and the square described on AB as AB^.
This notation, together with the signs 4- and -, will be employed
in the additional matter appended to this book ; tut it is not
adrniUed into Euclid's text because it ia desirable in the first instance
to emphasize the distinction between geometrical magnitudes them-
selves and the nuTrurical equivalents by wlijch they may be expresa«d
arithmetically.

H.B.E. 1

130

EUCLID'S ELEMENTS.

Proposition 1. Theorem.

If there are two straight lineSy one of which is divided into
any number of parts, the rectangle contained by the two straight
lines is equal to the sum of the rectangles contained by the vmr
divided straight line and the several parts of the divided line,

A D B

L H

Let P and AB be two straight lines, and let AB be
divided into any number of parts AC, CD, DB.

Then shall the rectangle contained by P, AB be equal to the
sum of the rectangles contained by P, AC, by P, CD, and by
P, DB.

Construction. From A draw AF perp. to AB; I. 11.

and make AG equal to P. I. 3.

Through G draw GH par^ to AB ; L 31.

and through C, D, B draw CK, DL, BH par* to AG.

Proof. Now the fig. AH is made up of the figs. AK, CL,
DH, and is therefore equal to their sum ;

and of these,

the fig. AH is the rectangle contained by P, AB ;

for it is contained by AG, AB ; and AG = P :

and the fig. AK is the rectangle contained by P, AC ;

for it is contained by AG, AC ; and AG = P :

also the fig. CL is the rectangle contained by P, CD ;

for it is contained by CK, CD ;

and CK = the opp. side AG, and AG = P. I. 34.

Similarly the fig. DH is the rectangle contained by P, DB.

.*. the rectangle contained by P, AB is equal to the
sum of the rectangles contained by P, AC, by P, CD, and
by P, DB. Q.E.D.

BOOK II. PROP. 1. 131

CORRESPONDING ALGEBRAICAL FORMULA.

In accordance with the principles explained on page 129, the
result of this proposition may be written thus :

P.AB=P. AC + P.CD + P. DB.

Now if the line P contains p units of length, and if AC, CD, DB
contain a, h, c units respectively,

then AB=a + 6 + c;
hence the statement

P.AB = P.AC + P.CD + P.DB
becomes p{a-i-h-i-c)=pa+pb-i-pc.

[Note. It must be understood that the rule given on page 129,
for expressing the area of a rectangle as the product of the lengths
of two adjacent sides, implies that those sides are coxmnensurable,
that is, that they can be expressed exactly in terms of some common
unit.

This however is not always the case. Two straight lines may be
so related that it is impossible to divide either of them into equal
parts, of which the other contains an exact number. Such lines are
said to be incommensurable. Hence if the adjacent sides of a rect-
angle are incommensurable, we cannot choose any linear unit in
terms of which these sides maybe exactly exjpreased ; and thus it
will be impossible to subdivide the rectangle into squares of unit
area, as illustrated in the figure of page 129. We do not here pro-
pose to enter further into the subject of incommensurable quantities :
it is sufficient to point out that further kno^edge of them will
convince the student that the area of a rectangle may be expressed
to any required degree of accuracy by the product of the lengths of
two adjacent sides, whether those lengths are commensurable or
not.]

132

EUCLID*S ELEMENTS.

Proposition 2. Theorem.

If a straight line is divided into any two parts, the squa/re
on the whole line is equal to the sum of the rectangles contained
by the whole lin^ and each of the pa/rts.

Let the straight line AB be divided at C into the two
parts AC, CB.

Then shall the square on AB he equal to the svm of the
rectangles contained by AB, AC, and by AB, BC.

Construction. On AB describe the square ADEB. I. 46.
Through C draw CF par^ to AD. I. 31.

Proof. Now the fig. AE is made up of the figs. AF, CE :

and of these,

the fig. AE is the sq. on AB : Constr,

and the fig. AF is the rectangle contained by AB, AC ;
for it is contained by AD, AC ; and AD = AB :

also the fig. CE is the rectangle contained by AB, BC ;
for it is contained by BE, BC ; and BE = AB.

.-. the sq. on AB = the sum of the rectangles contained
by AB, AC, and by AB, BC. Q.KD.

CORRESPONDING ALGEBRAICAL FORMULA.

The result of this proposition may be written

AB2=AB.AC+AB.BC.
Let AC contain a units of length, and let CB contain h units,

then AB = a + 6 units ;
and we have {a + hf={a + h)a + {a+h)h^

book ii. prop. 3.
Proposition 3. Theorem.

133

If a straight line is divided into any two parts, the rectangle
contained by the whole and one of the parts is equal to the
square on that part together with the rectangle contained by the
two parts,

A C B

Let the straight line AB be divided at C into the two
parts AC, CB.

Then shall the rectangle contained by AB, AC be equal to the
square on AC together with the rectangle contained by AC, CB.

Construction. On AC describe the square AFDC. I. 46.
Through B draw BE par^ to AF, meeting FD produced in E.

I. 31.

Proof. Now the fig. AE is made up of the figs. AD, CE ;

and of these,

the fig. AE is the rectangle contained by AB, AC ;

for AF = AC ;
and the fig. AD is the sq. on AC ; Constr,

also the fig. CE is the rectangle contained by AC, CB ;

for CD = AC.

.*. the rectangle contained by AB^ AC is equal to the
sq. on AC together with the rectangle contained by AC, CB.

Q.E.D.

CORRESPONDING ALGEBRAICAL FORMULA.

This result may be written AB . AC=AC2+AC . CB.

Let AC, CB contain a and b units of length respectively,

then AB =a + b units ;

and we have {a + h)a=a^-i-db.

Note. It should be observed that Props. 2 and 3 are specicU cases
of Prop. 1.

134

EUCLID S ELEMENTS.

Proposition 4. Theorem.

If a straight line is divided into any two parts, the square on
the whole line is equal to the sum of the squares on the two parts
together with twice the rectangle contained by the two parts.

Let the straight line AB be divided at C into the two
parts AC, CB.

ITien shall the sq. on AB he equal to the sum of the sqq, on
AC, CB, together with twice the rect. AC, CB.

Construction. On AB describe the square ADEB. L 46.

Join BD.
Through C draw CF par* to BE, meeting BD in G. I. 31.
Through G draw HGK par* to AB.

It is first required to shew that the fig. CK is the
sq. on CB.

Proof. Because CF and AD are par*, and BD meets them,
.-. the ext. angle CGB = the int. opp. angle ADB. L 29.
And since AB = AD, being sides of a square ;

.'. the angle ADB = the angle ABD; I. 5.

.-. the angle CGB = the angle CBG.

.-. CB = CG. L 6.

And the opp. sides of the par" CK are equal ; I. 34.

.'. the fig. CK is equilateral;

also the angle CBK is a right angle ; Def 30.

.'. CK is a square, and it is described on CB. I. 46, Cor,

Similarly, the fig. HF is the sq. on HG, that is, the
sq. on AC ;

for HG = the opp. side AC. L 34.

BOOK II. PROP. 4. 135

Again, the complement AG = the complement GE ; I. 43.
and the fig. AG = the rect. AC, CB ; for CG = CB.
.-. the two figs. AG, GE = twice the rect. AC, CB.

* Now the sq. on AB = the fig. AE

= the figs. HF, CK, AG, GE

= the sqq. on AC, CB together with

twice the rect. AC, CB.
.*. the sq. on AB = the sum of the sqq. on AC, CB with

twice the rect. AC, CB. q.e.d.

Corollary 1. Parallelograms about the diagonals of a
square are themselves squares.

Corollary 2. If a straight line is bisected, the square on
the whole line is four tim>es the square on half the line,

* For the purpose of oral work, this step of the proof
may conveniently be arranged as follows :

Now the sq. on AB is equal to the fig. AE,

that is, to the figs. HF, CK, AG, GE;
that is, to the sqq. on AC, CB together
with twice the rect. AC, CB.

CORRESPONDING ALGEBRAICAL FORMULA.

The result of this important Proposition may be written thus

AB2=AC2+CB2+2AC . CB.
Let AC=a, and CB=6;

then AB=a + 6 ;
hence the statement AB2=AC2 + CB2+2AC . CB
becomes {a + bf=a^ + h^+2ab.

136

EUCLID'S ELEMENTS.

Proposition 5. Theorem.

If a straight line is divided egmlly and also v/nequally^
the rectangle contained by the unequal parts, together with the
square on the line between the points of section, is equal to the
square cfn, half the lin^e,

A P Q B

C ED

Let the straight line AB be divided equally at P, and
unequally at Q.

Then the rect. AQ, QB together with the sq, on PQ shall be
equal to the sq, on PB.

Construction. On PB describe the square PCDB. L 46.

Join BC.
Through Q draw QE par^ to BD, cutting BC in F. L 31.
Through F draw LFHG par^ to AB.
Through A draw AG par^ to BD.
Proof.

Now the complement PF = the complement FD ; L 43.

to each add the fig. QL ;
then the fig. PL = the fig. QD.
But the fig. PL = the fig. AH, for they are par™* on equal
bases and between the same par^ ; i. 36.

.-. the fig. AH = the fig. QD.

To each add the fig. PF ;

then the fig. AF = the gnomon PLE.

Now the fig. AF is the rect. AQ, QB ; for QF = QB ;
.-. the rect. AQ, QB = the gnomon PLE.
To each add the sq. on PQ, that is, the fig. HE ; ii. 4.
then the rect. AQ, QB with the sq. on PQ

= the gnomon PLE with the fig. HE
= the whole fig. PD,
which is the sq. on PB.

BOOK n. PROP. 5. 137

That is, the rect. AQ, QB together with the square on
PQ is equal to the sq. on PB. q.e.d.

Corollary. From this Proposition it follows that the
difference of the squwres on two straight lines is equal to the
rectangle contained by their sum and difference.

For let X and Y be the given a P Q B

St. lines, of which X is the greater. ' '

Draw AP equal to X, and pro- ^
duce it to B, making PB equal to

AP, that is to X. Y

From PB cut off PQ equal to Y.

Then AQ is equal to the sum of X and Y,

and QB is equal to the difference of X and Y.

Now because AB is divided equally at P and unequally at Q,

.*. the rect. AQ, QB with sq. on PQ=the sq. on PB ; ii. 5.

that is, the difference of the sqq. on PB, PQ=the rect. AQ, QB.

or, the difference of the sqq. on A and Y=the rectangle contained

by the sum and the difference of X and Y.

CORRESPONDING ALGEBRAICAL FORMULA.

This result may be written

AQ.QB + PQ2=PB2.
Let AB=2a ; and let PQ=6 ;

then AP and PB each = a.
AlsoAQ=a + 6; andQB=a-&.
Hence the statement AQ . QB + PQ2= PB*
becomes (a + 6) (a - 6) + 6* = a*,

or (a + 6)(a-6)=a2_62.

EXERCISE.

In the above figure shew ihcU AP is half the sum of AQ and QB ;
and that PQ is half their difference.

138

EUCLID'S ELEMENTS.

Proposition 6. Theorem.

If a straight line is bisected and produced to any point, the
rectangle contained by the whole line thus produced and the part
of it produced, together with the square on half the line bisected,
is equal to the square on the straight line rnade up of the half
amd the part produced.

A P B Q

Let the straight line AB be bisected at P, and produced
to Q.

Then the rect AQ, QB together with the sq, on PB shall be
equal to the sq, on PQ.

Construction. On PQ describe the square PCDQ. i. 46.

Join QC.
Through B draw BE par* to QD, meeting QC in F. L 31.
Through F draw LFHG par* to AQ.
Through A draw AG par* to QD.

Proof. Now the complement PF = the complement FD. I. 43.

But the fig. PF = the fig. AH ; for they are par*"' on

equal bases and between the same par**. i. 36.

.-. the hg, AH = the fig. FD.

To each add the fig. PL ;

then the fig. AL = the gnomon PLE.

Now the fig. AL is the rect. AQ, QB ; for QL = QB ;

. '. the rect. AQ, QB = the gnomon PLE.

To each add the sq. on PB, that is, the fig. HE ;

then the rect. AQ, QB with the sq. on PB

= the gnomon PLE with the fig. HE
= the whole fig. PD,
which is the square on PQ.
That is, the rect. AQ, QB together with the sq. on PB is
equal to the sq. on PQ. Q.E.D.

BOOK II. PROP. 6. 139

CORRESPONDING ALGEBRAICAL FORMULA.

This result may be written

AQ.QB+PB2=PQ2.

Let AB=2a ; and let PQ=6 ;

then AP and PB each = a.
AlsoAQ=a + 6; andQB=&-a.

Hence the statement AQ . QB + PB2= PQ^
becomes (a + 6)(6-a) + a^=6^

or {b + a)(b-a)=b^-a^.

Definition. If a point X is taken in a straight line AB, or in
AB produced, the distances of

the point of section from the A ^ B

extremities of AB are said to be
the segments into which AB is
divided at X. a d y

In the former case AB is '^ ^ ^

divided internally, in the latter case externally.

Thus in each of the annexed figures, the segments into which
AB is divided at X are the lines AX and XB.

This definition enables us to include Props. 5 and 6 in a single
Enunciation.

If a stratgJU line is bisectedf and also divided {internally or ex-
ternally) into two unequal segments^ the rectangle contained by the
unequal segments is equal to the difference of the squares on half the
line, atid on the line between the points of section.

EXERCISE.

Shew that the EInunciations of Props. 5 and 6 may take the
following form :

llie rectangle contained by two straight lines is equal to the differ-
ence of the squares on hcdf their sum and on haZf their difference,

[See Ex., p. 137.]

EUCLID'S ELEMENTS.

Proposition 7. Theorem.

If a straight line is divided into any two parts, the sum of
the squares on the whole line and on one of the pa/rts is equal to
tvjice the rectangle contained by the whole and that part, together
vnth the square on the other part.

C B

D F E

Let the straight line AB be divided at C into the two
parts AC, CB.

Then shall the sum of the sqq. on AB, BC be equal to twice
the rect, AB, BC together vnth the sq, on AC.

Construction. On AB describe the square ADEB. I. 46.

Join BD.
Through C draw CF par^ to BE, meeting BD in G. I. 31.
Through G draw HGK par^ to AB.

Proof. Now the complement AG = the complement GE ; I. 43.

to each add the fig. CK :

then the fig. AK = the fig. CE.

But the fig. AK is the rect. AB, BC ; for BK = BC ;

.'. the two figs AK, CE = twice the rect. AB, BC.

But the two figs. AK, CE make up the gnomon AKF and the

fig. CK :
.* . the gnomon AKF with the fig. CK = twice the rect. AB, BC.

To each add the fig. H F, which is the sq. on AC :
then the gnomon AKF with the figs. CK, HF

= twice the rect. AB, BC with the sq. on AC.
But the gnomon AKF with the figs. CK, HF make up the

figs. AE, CK, that is to say, the sqq. on AB, BC ;
.*. the sqq. on AB, BC = twice the rect. AB, BC with the
sq. on AC. Q.E.D.

BOOK II. PROP. 7. 141

CORRESPONDING ALGEBRAICAL FORMULA.

The result of this proposition may be written

AB2 + BC2=2AB . BC + AC2.
Let AB=a, and BC=6 ; then AC=a-6.
Hence the statement

AB2+BC2=2AB. BC + ACa
becomes a^+h^= 2ah + (a- b)^,

or {a-h)^=a^-2ab + b^.

Comparing this result with that obtained from Prop. 4, we see
that

(i) 7%e sqtuire on the sum of two straight lines is greater than the
sum of the squares on those lines by twice the rectangle contained by
them. [Prop. 4.]

(ii) The square on the difference of two straight lines is less than
the sum of the squares on those lines by twice the rectangle contained by
them, [Prop. 7.]

ALTERNATIVE PROOFS OF PROPOSITIONS 4, 5, 6, 7.

The following alternative proofs are recommended for purposes
of revision, as affording useful exercise on the enunciations of pre-
ceding propositions, and illustrating the way in which many examples
on Book II. may be solved. The beginner however should not adopt
these proofs until he has thoroughly mastered those given in the
text, where the rectangles and squares are actually represented in
the diagrams.

Proposition 4.

Let the straight line AB be divided at C into two parts AC, CB.
Then shall tM. sq. on AB be equal to the sum of the sqq. on AC, CB
with twice the red. AC, CB.

A C B

Now the sq. on AB=the rect. AB, AC with the rect. AB, CB. ii. 2.

But the rect. AB, AC = the sq. on AC with the rect. AC, CB ; ii. 3.

and the rect. AB, CB=the sq. on CB with the rect. AC, CB. ii. 3.

Hence the sq. on AB=the sum of the sqq. on AC, CB with twice

the rect. AC, CB.

142 EUCLID*S ELEMENTS.

Proposition 6.

Let the straight line AB be divided equally at P, and unequally
at Q.

Then ahail the red, AQ, QB vnth the sq. on PQ he equal to the
sq. on PB.

A P 9 B

Now the rect. AQ, QB = the rect. AP, QB with the rect. PQ, QB, n. 1.

=therect. PB, QB with the rect. PQ, QB.
But the rect. PB, QB=the sq. on QB with the rect. PQ,QB ; ii. 3.
.'. the rect. AQ, QB=the sq. on QB with tivice the rect. PQ, QB.
To each of these equals add the sq. on PQ.
Then the rect. AQ, QB with the sq. on PQ

=the sqq. on PQ, QB with twice the rect. PQ, QB
=the sq. on PB. ii. 4.

Proposition 6.

Let the straight line AB be bisected at P, and produced to Q.
Then shcUl the rect, AQ, QB with the sq. on PB he equcU to the
sq, on PQ.

A P |_Q

Now the rect. AQ, QB=the rect. AP, BQ with the rect. PQ, BQ

II. 1.
= the rect. PB, BQ with the rect. PQ, BQ.
But the rect. PQ, BQ=the sq. on BQ with the rect. PB, BQ. ii. 3.
.'. the rect. AQ, QB = the sq. on BQ with twice the rect. PB, BG^
To each of these equals add the sq. on PB.
Then the rect. AQ, QB with the sq. on PB

=the sqq. on PB, BQ with twice the rect. PB, BQ
= the sq. on PQ. ii. 4.

Proposition 7.

Let the straight line AB be divided at any point C.
Then shall the sum of the sqq, on AB, BC he equal to twice the
rect, AB, BC with the sq, on AC.

A 9 B

Now the sq. on AB=the sqq. on AC, CB with twice the rect. AC, CB.

II. 4.
To each of these equals add the sq. on BC.
Then the sqq. on AB, BC=the sq. on AC with tioice the sq. on BC

and tmce the rect. AC, CB.
But twice the sq. on BC with twice the rect. AC, CB

= twice the rect. AB, BC. n. 3.
.'. the sqq. on AB, BC=the sq. on AC with twice the rect. AB, BC.

BOOK II. PROP. 8.

143

Ohs, The following proposition being little used, we merely give
the figure and the leading points of Euclid's proof.

Proposition 8. Theobbm.

If a straight line is diinded into any two parts, four times the
rectangle contained by the whole line and owe of the parts, together
with the square on the other part, is equal to the sqv/ire on the straight
line which is made up of the whole and the first named part.

Let AB be divided at C.
Produce AB to D, making BD equal to
BC. M

TJien shall four times the red, AB, BC X
with the sq. on AC=t?ie sq. on AD.

On AD describe the square AEFD ; and
complete the construction as indicated in
the figure.

Euclid then proves (i) that the figs. CK,
BN, QR, KO are all equal :

L F

(ii) that the figs. AG, MP, PL, RF are all equal.

Hence the eight figures named above are together four times the
sum of the figs. AG, CK; that is, four times the fig. AK; that is,
four times the rect. AB, BC.

But the whole fig. AF, namely the sq. on AD, is made up of these
eight figures, together with the fig. XH, which is the sq. on AC :

hence the sq. on AD = four times the rect. AB, BC, together with
the sq. on AC. q.e.d.

C B D

The accompanying figure will suggest a
less cumbrous proof, which we leave as an
Exercise to the student.

CORRESPONDING ALGEBRAICAL FORMULA.

The result of this proposition may be written

4AB.BC + AC2=AD2.
Let AB=a, and BC = 6; then AC=a-6, and AD=a + &.
Hence we have 4ab + (a - 6)^ ={a + hf ;
or (a + 6)2-(a-6)2=4a6.

144

EUCLID'S ELEMENTS.

Proposition 9. Theorem. [Euclid's Proof.]

If a straight line is divided equally and also' unequally, the
sum of the squares on the two unequal parts is twice the sum of
the squares on half the line and on the line between the points of
section.

Let the straight line AB be divided equally at P, and
unequally at Q.

Then shall the sum of the sqq. on AQ, QB be twice the sum
of the sqq, on AP, PQ. •

Construction. At P draw PC at rt. angles to AB; I. 11.
and make PC equal to AP or PB. i. 3.

Join AC, BC.
Through Q draw QD par^ to PC ; I. 31.

and through D draw DE par^ to AB.

Join AD.

Proof. Then since PA= PC, Constr,

,\ the angle PAC = the angle PCA. L 5.

And since, in the triangle APC, the angle APC is a rt.

angle, Constr,

.-. the sum of the angles PAC, PCA is a rt. angle : L 32.

hence each of the angles PAC, PCA is half a rt. angle.

So also, each of the angles PBC, PCB is half a rt. angle.

.'. the whole angle ACB is a rt. angle.

Again, the ext. angle CED = the int. opp. angle CPB; I. 29.

.'. the angle CED is a rt. angle :
and the angle ECD is half a rt. angle. Proved.
.*. the remaining angle EDC is half a rt. angle ; i, 32.
.-. the angle ECD = the angle EDC;

.-. EC = ED. I. 6.

BOOK II. PROP. 9. 145

Again, the ext. angle DQB = the int. opp. angle CPB ; I. 29.

.*. the angle DQB is a rt. angle.
And the angle QBD is half a rt. angle ; Proved.
.*. the remaining angle QDB is half a rt. angle ; i. 32.
.*. the angle QBD = the angle QDB ;
.-. QD=:QB.

Now the sq. on AP = the sq. on PC ; for AP = PC. Constr,
And since the angle APC is a rt. angle,

.*. the sq. on AC = the sum of the sqq. on AP, PC ; i. 47.
.*. the sq. on AC is twice the sq. on AP.

Similarly, the sq. on CD is twice the sq. on ED, that is, twice

the sq. on the opp. side PQ. I. 34.

Now the sqq. on AQ, QB = the sqq. on AQ, QD Proved.

= the sq. on AD, for AQD is a rt.

angle ; I. 47.

= the sum of the sqq. on AC, CD,

for ACD is a rt. angle ; I. 47.

= twice the sq. on AP with twice

the sq. on PQ. Proved,

That is,

the sum of the sqq. on AQ, QB = twice the sum of the sqq.

on AP, PQ. Q.E.D.

CORRESPONDING ALGEBRAICAL FORMULA.

The result of this proposition may be written

AQ2 + QB2=2(AP+PQ2).

Let AB=2a ; and PQ=6 ;

then AP and PB each = a.
Al8oAQ=a + &; andQB=a-&.
Hence the statement

AQ2 + QB2 = 2 ( AP + QP)
becomes (a + h)^ + {a-h)^=2{a^ + h%

[Note. For alternative proofs of this proposition, see page 148.]

H.S.E. K

146 euclid's elements.

Proposition 10. Theorem. [Euclid's Proof.]

If a straight line is bisected and ^produced to any point, the
sum of the squares on the whole line thus produced, and on the
'part produced, is twice the sum of the squares on half the line
bisected and on the line made up of the half and the part
produced.

E D

Let the st. line AB be bisected at P, and produced to Q.
Then shall the sum of the sqq. on AQ, QB be twice the sum
of the sqq, on AP, PQ.

Construction. At P draw PC at right angles to AB ; I. 11.

and make PC equal to PA or PB. I. 3.

Join AC, BC.

Through Q draw QD par^ to PC, to meet CB produced

in D ; L 31.

and through D draw DE par^ to AB, to meet CP produced

in E.

Join AD.

Proof. Then since PA=PC, Constr,

.*. the angle PAC = the angle PCA. i, 5.

And since, in the triangle APC, the an^e APC is a rt. angle,
.'. the sum of the angles PAC, PCA is a rt. angle, i. 32.
Hence each of the angles PAC, PCA is half a rt. angle.
So also, each of the angles PBC, PCB is half a rt. angle.
.•. the whole angle AC B is a rt. angle.

Again, the ext. angle CPB = the int. opp. angle CED : I. 29.

.*. the angle CED is a rt. angle :
and the angle ECD is half a rt. angle ; Proved,
.', the remaining angle EDC is half a rt. angle. L 32.
.*. the angle ECD = the angle EDC ;

.-. EC = ED. L 6.

BOOK II. PROP. 10. 147

Agaii\, the angle DQB = the alt. angle CPB; i. 29.
.*. the angle DQB is a rt. angle.
Also the angle QBD = the vert. opp. angle CBP: I. 15.

that is, the angle QBD is half a rt. angle.
.*. the remaining angle QDB is half a rt. angle : L 32.
.'. the angle QBD = the angle QDB ;

.*. QB = QD. I. 6.

Now the sq. on AP = the sq. on PC ; for AP= PC. Constr.
And since the angle APC is a rt. angle,
.*. the sq. on AC = the sum of the sqq. on AP, PC ; I. 47.
.*. the sq. on AC is twice the sq. on AP.

Similarly, the sq. on CD is twice the sq. on ED, that is, twice
the sq. on the opp. side PQ. i. 34.

Now the sqq. on AQ, QB = the sqq. on AQ, QD Proved.

= the sq. on AD, for AQD is a rt.

angle; I. 47.

= the sum of the sqq. on AC, CD,

for AC D is a rt. angle ; I. 47.

= twice the sq. on AP with twice

the sq. on PQ. Proved.

That is,

the sum of the sqq. on AQ, QB is twice the sum of the sqq.

on AP, PQ. Q.E.D.

CORRESPONDING ALGEBRAICAL FORMULA.

The result of this proposition may he written

AQ2 + QB2=2(AP+PQ2).
Let AB=2a ; and PQ=6 ;

then AP and PB each = a.
Also AQ=a + 6 ; and QB=6-a.
Hence we have

{a + hf + {h-af=2{a^+h^).

[Note. For alternative proofs of this proposition, see page 149.]

148 EUCLID'S ELEMENTS.

Proposition 9. [Alternative Proof.]

If a sl/raight line is divided equally and also uneqiiaUy, the
sum of the squares on the two unequal parts is twice the sum of
the squares on half the line and on the line between the points of
section,

A P Q B

Let the straight line AB be divided equally at P and
unequally at Q.

Then shall the sum of the sqq. on AQ, QB be tvnce the sum
of the sqq. on AP, PQ.

Proof.
The sq. on AQ = the sum of the sqq. on AP, PQ with twice

the rect. AP, PQ ii. 4.

= the sum of the sqq. on AP, PQ with twice
the rect. PB, PQ; for PB = AP.

To each of these equals add the sq. on QB.
Then the sqq. on AQ, 08 = the sum of the sqq. on AP, PQ

with twice the rect. PB, PQ
and the sq. on QB.

But twice the rect. PB, PQ and the sq. on QB

= the sum of the sqq. on PB, PQ. n. 7.

.-. the sqq. on AQ, QB = the sum of the sqq. on AP, PQ with

the sum of the sqq. on PB, PQ
= twice the sum of the sqq. on AP, PQ.

Q.E.D.

Note. The following concise proof, obtained from ii. 4 and
II. 5, is useful as an exercise, but it is hardly admissible as a formal
demonstration owing to its algebraical use of the negative sign.

We have AQ2+QB2=AB2-2AQ . QB ii. 4.

=4PB2-2AQ . QB ii. 4, Cor. 2.

=4PB2-2(PB2-PQa) n. 5.
=2PB2+2PQ3.

BOOK 11. PROP. 10, 149

Proposition 10. [Alternative Proof.]

If a straight line is bisected and p-oduced to any point, the
sum of the squares on the whole line thus produced and on the
part produced, is twice the sum of the squares on half the line
bisected and on the line made up of the half and the part
produced.

A P B Q

Let the straight line AB be bisected at P, and produced
toQ.

Then shall the sum of the sqq. on AQ, QB be twice the swm
of the sqq. on AP, PQ.

Proof.
The sq. on AQ = the sum of the sqq. on AP, PQ with twice

the rect. AP, PQ ii. 4.

= the sum of the sqq. on AP, PQ with twice
the rect. PB, PQ ; for PB = AP.

To each of these equals add the sq. on QB.
Then the sqq. on AQ, QB = the sum of the sqq. on AP, PQ

with twice the rect. PB, PQ
and the sq. on QB.

But twice the rect. PB, PQ and the sq. on QB

= the sum of the sqq. on PB, PQ. II. 7.

.*. the sqq. on AQ, QB = the sum of the sqq. on AP, PQ with

the sum of the sqq. on PB, PQ
= tunce the sum of the sqq. on AP, PQ.

Q.E.D.

Note. Another proof of this proposition, based on ii. 7 and
n. 6, is indicated by the following steps :

We have AQ2 + QB^ = 2AQ . QB + AB^ ii. 7.

=2AQ . QB+4PB2 n. 4, Cor. 2.

= 2 ( PQ2 - PB2) + 4PB3 II. 6.
=2PB2+2PQa.

150

Euclid's blbments.

Proposition 11. Problem.

To divide a given straight line into two parts, so that ihe
rectangle contained by the whole and one part may be equal
to the square on the other paii.

Let AB be the given straight line.
It is required to divide AB into two parts, so that the rectangle
contained by the whole and one part may be equal to the square
on the other part.

Ck>n8truction.

L 46.
L 10.

L 3.
L 46.

On AB describe the square ACDB.

Bisect AC at E.

Join EB.

Produce CA to F, making EF equal to EB.

On AF describe the square AFGH.

Then shall AB be divided at H, so that the rect, AB, BH is equal

to the sq. on AH.

Produce GH to meet CD in K.

Proof. Because CA is bisected at E, and produced to F,
. *. the rect. CF, FA with the sq. on EA = the sq. on EF ll. 6.

= the sq. on EB. Constr,
But the sq. on EB = the sum of the sqq. on EA, AB,

for the angle EAB is a rt. angle. I. 47.

.'. the rect. CF, FA with the sq. on EA = the sum of the
sqq. on EA, AB.

From these equals take the sq. on EA :
then the rect. CF, FA = the sq. on AB.

BOOK II. PROP. 11. 151

But the rect. CF, FA = the fig. FK; for FA= FG;
and the sq. on AB = the fig. AD. Constr,

.-. the fig. FK = thefig. AD.

From these equals take the common fig. AK ;

then the remaining fig. FH = the remaining fig. HD.

But the fig. HD = the rect. AB, BH ; for BD = AB;

and the fig. FH is the sq. on AH.

.•. the rect. AB, BH = the sq. on AH. Q.E.F.

Definition. A straight line is said to be divided in Medial
Section when the rectangle contained by the given line and one of
its segments is equal to the square on the other segment.

The student should observe that this division may be internal or
extemcU.

Thus if the straight line AB is divided internally at H, and ex-
ternally at H', so that

(!) AB.BH =AH2, lj' A H B

(ii) AB.BH'=AH'2, '

we shall in either case consider that AB is divided in medial section.
The case of internal section is alone given in Euclid ii. 11 ; but a
straight line may be divided externally in medial section by a similar
process. See Ex. 21, p. 160.

ALGEBRAICAL ILLUSTRATION.

It is required to find a point H in AB, or AB produced, such that

AB.BH =AH2.
Let AB contain a units of length, and let AH contain x units ;

then BH =a-aj:
and X must be such that a(a- x)=x^,
or 3t^ + ax-a^=0.

Thus the construction for dividing a straight line in medial section
corresponds to the solution of this quadratic equation, the two roots
of which indicate the intei-nal and external points of division.

EXERCISES.

In the figure of ii. 11, shew that
(i) if CH is produced to meet BF at L, CL is at right angles

toBF;
(ii) if BE and CH meet at O, AO is at right angles to CH.
(iii) the lines BG, DF, AK are parallel :
(iv) CF is divided in medial section at A.

152 EUCLID'S ELEliENTS.

Proposition 12. Theorem.

In an obtuse-omgUd triangle^ if a perpendictdcMr is dravm
from either of the acute angles to the opposite side produceij the
square on the side subtending the obtuse angle is greater than the
sum of the squares on the sides containing the obtuse anghy
by twice the rectangle contained by the side on which, when
produced, the perpendicular falls, and the line intercepted wiihout
the triangle, bettoeen the perpendicular and the obtuse angle.

Let ABC be an obtuse-angled triangle, having the obtuse
angle at C; and let AD be drawn from A perp. to the
opp. side BC produced. ^

Then shall the sq, on AB be greater tham, the sum of the
sqq. on BC, CA, by twice the rect BC, CD.

Proof. Because BD is divided into two parts at C,
.*. the sq. on BD = the sum of the sqq. on BC, CD, with
twice the rect. BC, CD. ii. 4,

To each of these equals add the sq. on DA.
Then the sqq. on BD, DA = the sum of the sqq. on BC, CD,
DA, with twice the rect. BC, CD.

But the sum of the sqq. on BD, DA = the sq. on AB,

for the angle at D is a rt. angle. i. 47.

Similarly, the sum of the sqq. on CD, DA = the sq. on CA.

.•. the sq. on AB = the sum of the sqq. on BC, CA, with
twice the rect. BC, CD.

That is, the sq. on AB is greater than the sum of the
sqq. on BC, CA by twice the rect. BC, CD. Q.E.D.

BOOK II. PROP. 12. 153

NOTE ON PROP. 12.

A general definition of the projection of one straiglit
line on another is given on page 105. The student's
attention is here called to a special case of projection which
will enable us to simplify the Enunciation of Proposition 12.

In the above diagram, CA is a given straight line drawn
from a point C in PQ; and from A a perpendicular AD is
drawn to PQ. In this case, CD is said to be the projection
of CA on PQ.

By applying this definition to the figure of Prop. 12, we
see that the statement

The sq. on ^B is greater than the sum of the sqq. on BC, CA
by twice the red, BC, CD

is the particular form of the following general Enunciation :

In an Muse^ngled triangle the square on the side opposite the
obtuse angle is greater than the sum of the squares on the sides
containing the obtuse angle by twice the rectangle contained by
one of those sides, and the projection of the other side upon it.

The Enunciation of Prop. 12 thus stated should be
carefully compared with that of Prop. 13.

154

EUCLID'S ELEMENTS.

Proposition 13. Theorem.

In every triangle, the square on the side svhtending an acute
angle is less than the sum of the squares on the sides containing
that angle, by twice the rectangle contained by either of these
sides, and the straight line intercepted between the perpendicular
let fall on it from the opposite angle, and the acute angle.

Fig. 2.

Let ABC be any triangle having the angle at C an acute
angle ; and let AD be the perp. drawn from A to the opp.
side BC.

Then shall the sq, on AB be less than the sum of the sqq. on
BC, CA, by tvnce the red, BC, CD.

Proof. Now AD may fall within the triangle ABC, as in
fig. 1, or without it, as in fig. 2.

■D fin ^g. 1, BC is divided into two parts at D,

ecause ^^^ ^^ ^^ pc is divided into two parts at B,

.*. in both cases
the sum of the sqq. on BC, CD = twice the rect. BC, CD with
the sq. on BD. ll. 7.

To each of these equals add the sq. on DA.
Then the sum of the sqq. on BC, CD, DA = twice the rect.

BC, CD with the sum of the sqq. on BD, DA.
But the sum of the sqq. on CD, DA = the sq. on CA, I. 47.

for the angle ADC is a rt. angle.
Similarly, the sum of the sqq. on BD, DA = the sq. on AB.

.'. the sum of the sqq. on BC, CA = twice the rect. BC, CD

with the sq. on AB.
That is, the sq. on AB is less than the sum of the sqq. on

BC, CA by twice the rect. BC, CD. Q.E.D.

BOOK II. PROP. 13. 155

Obs, If the perpendicular AD coincides with AB, that is, if
ABC is a right angle, then tivice the red, BC, CD becomes ttoice the
sq, an BC ; and it may be shewn that the proposition merely repeats
the result of i. 47.

NOTES ON PROP. 13.

(i) Remembering the definition of the projection of a
straight line given on p. 153, we may enunciate Prop. 13
as foUows ;

In every triangUy the square on the side suhtending an acute
angle is less than the sum of the squares on the sides containing
that angkj by tunce the rectangle contained by one of these sides
and the projection of the other side upon it

(ii) Comparing the Enunciations of ii. 12, i. 47, li. 13,
we see that in the triangle ABC,

if the angle ACB is obtuse^ we have by il. 12,
AB2=BC2 + CA2+2BC.CD;

if the angle ACB is a right angle, we have by I. 47,
AB2 = BC2 + CA2 ;

if the angle ACB is acute, we have by ii. 13,
AB2= BC2 + CA2 - 2BC . CD.

These results may be collected as follows :

The square on a side of a triangle is greater than, equal to, or
less than the sum of the squares on the other sides, according as
the angle opposite to the first is obtuse, a right angle, or acute,

EXERCISES ON II. 12 AND 13.

1. If from one of the base angles of an isosceles triangle a per-
pendicular is drawn to the opposite side, then twice the rectangle
contained by that side and the segment adjacent to the base is equal
to the square on the base.

2. If one angle of a triangle is one-third of two right angles,
shew that the square on the opposite side is less than the sum of the
squares on the sides forming that angle, by the rectangle contained
by these two sides. [See Ex. 10, p. 109.]

3. If one angle of a triangle is two-thirds of two right angles,
shew that the square on the opposite side is greater than the sum of
the SQuares on the sides formmg that angle, by the rectangle con-
tainea by these sides. [See Ex. 10, p. 109.]

156

EUCLID'S ELEMENTS.

Proposition 14. Problem.

To describe a square that shM be equal to a given rectilineal
figure.

Let A be the given rectilineal figura
It is required to describe a square equal to A.

Construction. Describe a par"" BCDE equal to the fig. A,
and having the angle CBE a right angle. i. 45.

Then if BC = BE, the fig, BD is a square; and what was

required is done.
But if not, produce BE to F, making EF equal to ED ; I. 3.

and bisect BF at G. I. 10.

With centre G, and radius GF, describe the semicircle BHF;
produce DE to meet the semicircle at H.
Then shall the sq. on EH be equal to the given fig. A.

Join GH.

Proof. Because BF is divided equally at G and unequally
at E,

/. the rect. BE, EF with the sq. on GE = the sq. on GF IL 5.

= the sq. on GH.

But the sq. on GH = the sum of the sqq. on GE, EH ;

for the angle H EG is a rt. angle. L 47.

.". the rect. BE, EF with the sq. on GE = the sum of the
sqq. on GE, EH.

From these equals take the sq. on GE :
then the rect. BE, EF = the sq. on HE.
But the rect. BE, EF = the fig. BD ; for EF = ED ; Constr,
and the fig. BD = the given fig. A. Constr,

.*. the sq. on EH = the given fig. A. Q.E.F.

QUESTIONS FOR REVISION. 157

QUESTIONS FOR REVISION ON BOOK II.

1. Explain the phrase, the rectangle contained by AB, CD ; and
shew by superposition that if AB = PQ, and CD = RS, then the
rectangle contained by AS, CD = the rectangle contained by PQ, RS.

2. Shew that Prop. 2 is a special case of Prop. 1, explaining
tinder what conditions Prop. 1 becomes identical with Prop. 2.

3. What must be the relation between the divided and un-
divided lines in the enunciation of Prop. 1 in order to give the
result proved in Prop. 3 ?

4. Define the segments into which a straight line is divided at a
point in such a way as to be applicable to the case when the dividing
point is in the given line produced.

Hence frame a statement which includes the enunciations of both
II. 5 and ii. 6, and find the algebraical formulae corresponding to
these enunciations.

Also combine in a single enunciation the results of i. 9 and ii. 10.

6. Compare the results proved in Propositions 4 and 7 by finding
the algebraical formulae corresponding to their enunciations.

6. The difference of the squares on two straight lines is equal to
the rectangle contained by their sum and difference. Deduce this
theorem from Prop. 5.

7. Define the projection of one straight line on another.

How may the enunciations of ii. 12 and ii. 13 be simplified by
means of this definition ?

8. In the figure of Proposition 14,

(i) If BE = 8 inches, and ED=2 inches, find the length of EH.

(ii) If BE = 12*5 inches, and EH=2'5 inches, find the length
of ED.

(iii) If BE =9 inches, and EH =3 inches, find the length of GH.

9. When is a straight line said to be divided in medial section ?

If a straight line 8 inches in length is divided internally in
medial section, shew that the lengths of the segments are approxi-
mately 4*9 inches and 3*1 inches.

[Frame a quadratic equation as explained on page 151, and solve.]

158 EUCLID'S ELEMENTS.

THEOREMS AND EXAMPLES ON BOOK 11.

ON II. 4 AND 7.

1. Shew hy ii. 4 that the square cm, a straight line is four times
the square on half the line,

[This result is constantly used in solving examples on Book n.,
especially those which follow from ii. 12 and 13.]

2. If a straight line is divided into any three parts, the square
on the whole line is equal to the sum of the squares on the three
parts together with twice the rectangles contained by each pair of
these parts.

Shew that the algebraical formula corresponding to this theorem

*s (a + 6 + c)2=o2 + 62+c2 + 26c+2co+2a6.

3. In a right-angled triangle, if a perpendicular is dravm from
the right angle to the hypotenuse, the square on this perpendicular is
equal to the rectangle contained by the segments of the hypotenuse,

4. In an isosceles triangle, if a perpendicular is drawn from one
of the angles at the base to the opposite side, shew that the square
on the perpendicular is equal to twice the rectangle contained by the
segments of that side together with the square on the segment
adjacent to the base.

5. Any rectangle is half the rectangle contained by the diagonals
of the squares described upon its two sides.

6. In any triangle if a perpendicular is drawn from the vertical
angle to the base, the sum of the squares on the sides forming that
angle, together with twice the rectangle contained by the segments
of the base, is equal to the square on the base together with twice
the square on the perpendicular.

ON II. 5 AND 6.

Ohs. The student is reminded that these important propositions
are both included in the following enunciation :

The difference of the squares on two straight lines is equal to the
rectangle contained hy their sum and difference. [See Cor., p. 137].

7. In a right-angled triangle the square on one of the sides form-
ing the right angle is equal to the rectangle contained by the sum and
dinerence of the hypotenuse and the other side. [i. 47 and ii. 5, Cor.]

THEOREMS AND EXAMPLES ON BOOK II. 159

8.^ T?ie difference of the aqua/res on two sides of a iriangle is equal
to tvjice the rectangle contained by the haSe and the intercept between
the middle point of the base and the foot of the perpendicular draum
from the vertical angle to the base.

Let ABC be a triangle, and let P be the middle point of the base
BC : let AQ be drawn perp. to BC.

Then shall AB^ - AC2=2BC . PQ.

First, let AQ fall within the triangle.

Now AB2 = BQ2 + QA2^ i. 47.

alsoAC2=QC2+QA2,

.-. AB2-AC2=BQ2-QC2 Ax. 3.

= ( BQ + QC) ( BQ - QC) ii. 6, Cor.
= BC.2PQ Ex., p. 137.

= 2BC.PQ Q.E.D.

The case in which AQ falls outside the triangle presents no
difficulty.

9. The sqtiare on any straight line dravm from the vertex of an
isosceles triangle to the base is less than the square on one of the equal
sides by the rectangle contained by the segments of the base,

10. The square on any straight line drawn from the vertex of an
isosceles triangle to the base produced, is greater than the square on
one of the equal sides by the rectangle contained by the segments into
which the base is divided externally.

11. If a straight line is drawn through one of the angles of
an equilateral triangle to meet the opposite side produced, so that
the rectangle contamed by the segments of the base is equal to the
square on the side of the triangle ; shew that the square on the line
so drawn is double of the square on a side of the triangle.

12. If XY is drawn parallel to the base BC of an isosceles
triangle ABC, then the difference of the squares on BY and CY is
equal to the rectangle contained by BC, X Y. [See above, Ex. 8. ]

13. In a right-angled triangle, if a perpendicular is drawn from
the right angle to the hypotenuse, tlie square on either side forming
the right angle is equal to the rectangle contained by the hypotenuse
and the segment of it adjacent to that side.

160 EUCLID'S ELEMENTS.

ON II. 9 AND 10.

14. Deduce Prop. 9 from Props. 4 and 5, using also the theorem
that the square on a straight line is four times the square on half
the line.

15. Deduce Prop. 10 from Props. 7 and 6, using also the theorem
mentioned in the preceding Exercise.

16. If a straight line is divided equally, and also unequally, the
squares on the two unequal segments are together equal to twice the
rectangle contained by these segments together with four times the
square on the line between the points of section.

ON n. 11.

17. If o> straight line is divided intemcUly in medial section, and
from the greaier segment a part he taken equal to the less ; shew thai
the greater segment is also divided in medial section.

18. If a straight line is divided in medial section, the rectangle
contained by the sum and difference of the segments is equal to
the rectangle contained by the segments.

19. If AB is divided at H in medial section, and if X is the
middle point of the ^eater segment AH, shew that a triangle whose
sides are equal to AH, XH, BX respectively must be right-angled.

20. If a straight line AB is divided internally in medial section
at H, prove that the sum of the squares on AB, BH is three times
the square on AH.

21. Divide a straight line externally in medial section,

[Proceed as in n. 11, but instead of drawing EF, make EF' equal
to EB in the direction remote from A; and on AF' describe the
square AF'G'H' on the side remote from AB. Then AB will be
divided externally at H' as required.]

ON n. 12 AND 13.

22. In a triangle ABC the angles at B and C are acute : if E
and F are the feet of perpendiculars drawn from the opposite angles
to the sides AC, AB, shew that the square on BC is equal to the
sum of the rectangles AB, BF and AC, UE.

23. ABC is a triangle right-angled at C, and DE is drawn from
a point D in AC perpendicular to AB : shew that the rectangle
Ad, AE is equal to the rectangle AC, AD.

THEOREMS AND EXAMPLES ON BOOK II. 161

24. In any triangle the sum of the sqtuires on ttoo sides is eqtud to
ttoice the square on half the third side together with twice the square on
the median which bisects the third side.

Q C

Let ABC be a triangle, and AP the median bisecting the side BC.
Then shall AB2 + AC2=2BP+2AP.

Draw AQ perp. to BC.

Consider the case in which AQ falls within the triangle, but does
not coincide with AP.

Now of the angles APB, APC, one must be obtuse, and the other
acute : let APB be obtuse.

Then in the A APB, AB2= BP^ + AP + 2 BP . PQ, ii. 12.

Also in the A APC, AC2=CP + AP2-2CP . PQ. ii. 13.

ButCP=BP,
.-. CF=BF; and the rect. BP, PQ=the rect. CP, PQ,
Hence adding the above results,
AB2+AC2=2 . BP+2 . AP2. q.b.d.

The student will have no difficulty in adapting this proof to the
cases in which AQ falls without the triangle, or coincides with AP.

26. The sum of the sqtmres on the sides of a parallelogram is
equal to the sum of the squares on the diagonals.

26. In any quadrilateral the squares on the diagonals are to-
gether equal to twice the sum of the squares on the straight lines
joining the middle points of opposite sides. [See Ex. 9, p. 105. ]

27. If from any point within a rectangle straight lines are drawn
to the angular points, the sum of the squares on one pair of the lines
drawn to opposite angles is equal to the sum of the squares on the
other pair.

28. The sum of the squares on the sides of a quadrilateral is
greater than the sum of the squares on its diagonals by four times
the square on the straight line which joins the middle points of the
diagonals.

29. O is the middle point of a given straight line AB, and from
O as centre, any circle is described : if P be any point on its circum-
ference, shew that the sum of the squares on AP, BP is constant.

162 EUCLID'S ELEMENTS.

30. Given the base of a triangle, and the sum of the squares on
the sides forming the vertical angle ; find the locus of the vertex.

3L ABC is an isosceles triangle in which AB and AC are equal.
AB is produced beyond the base to D, so that BD is equal to AB.
Shew that the square on CD is equal to the square on Ab together
with twice the square on BC.

32. In a right-angled triangle the sum of the squares on the
straight lines drawn from the right angle to the points of tri-
section of the hypotenuse is equal to five times the square on the
line between the points of trisection.

33. Three times the sum of the squares on the sid6s of a tri-
angle is equal to four times the sum of the squares on the medians.

34. ABC is a triangle, and O the point of intersection of its
medians : shew that

AB2 + BC2 + CA2=3(OA2 + OB2+OC2).

35. ABCD is a quadrilateral, and X the middle point of the
straight line joining the bisections of the diagonals ; with X as centre
any circle is described, and P is any point upon this circle : shew
that PA2 + PB2+ PC2+ PD2 is constant, being equal to

XA2+XB2+XC2 + XD2 + 4XP2.

36. The squares on the diagonals of a trapezium are together
equal to the sum of the squares on its two oblique sides, with twice
the rectangle contained by its parallel sides.

PROBLEMS.

37. Construct a rectangle equal to the difference of two squares.

38. Divide a given straight line into two parts so that the
rectangle contained by them may be equal to the square described
on a given straight line which is less than half the straight line to
be divided.

39. Given a square and one side of a rectangle which is equal
to the square, find the other side.

40. Produce a given straight line so that the rectangle contained
by the whole line thus produced and the part produced, may be
equal to the square on another given line.

41. Produce a given straight line so that the rectangle contained
by the whole line thus produced and the given line shall be equal to
the square on the part produced.

42. Divide a straight line AB into two parts at C, such that
the rectangle contained by BC and another line X may be equal to
the square on AC.

BOOK III.

Book III. deals with the properties of Circles.

For convenience of reference the following definitions
are repeated from Book I.

I. Def, 15. A circle is a plane figure
bounded by one line, which is called the
circiimference, and is such that all straight
lines drawn from a certain point within
the figure to the circumference are equal
to one another: this point is called the
centre of the circle.

Note. Circles which have the same centre are said to be
concentric.

I. Def, 16. A radius of a circle is a straight line drawn
from the centre to the circumference.

I. Def, 17. A diameter of a circle is a straight line
drawn through the centre, and terminated both ways by
the circumference.

I. Def 18. A semicircle is the figure bounded by a
diameter of a circle and the part of the circumference cut
off by the diameter.

Note. From these definitions we draw the following inferences :

(i) The distance of a point from the centre of a circle is less
than the radius, if the point is within the circumference : and the
distance of a point from the centre is greater than the radius, if the
point is without the circumference.

(ii) A point is within a circle if its distance from the centre is
less than the radius : and a point is without a circle if its distance
from the centre is greater than the radius.

(iii) Circles of equal radius are equal in all respects ; that is to
say, their areas and circumferences are equal.

(iv) A circle is divided by any diameter into two parts which
are equal in all respects.

164 EUCLID'S EliEMENTS.

Definitions to Book III.

1. An arc of a circle is any part of the circumference.

2. A chord of a circle is the straight line which joins
any two points on the circumference.

Note. From these definitions it may be seen
that a chord of a circle, which does not pass
through the centre, divides the circumference
into two unequal arcs ; of these,* the greater is
called the major arc, and the less the minor arc.
Thus the major arc is greater, and the minor arc
less than the semi-circumference.

The major and minor arcs, into which a cir-
cumference is divided by a chord, are said to be
conjugate to one another.

3. Chords of a circle are said to be
equidistant from the centre, when the
perpendiculars drawn to them from the
centre are equal :

and one chord is said to be farther from
the centre than another, when the per-
pendicular drawn to it from the centre is
greater than the perpendicular drawn to
the other.

4. A secant of a circle is a straight
line of indefinite length, which cuts the
circumference in two points.

5. A tangent to a circle is a straight
line which meets the circumference, but
being produced, does not cut it. Such a
line is said to touch the circle at a point ;
and the point is called the point of
contact.

DEFINITIONS.

165

Note. If a secant, which cuts a circle
at the points P and Q, gradually changes its
position in such a way that P remains fixed,
the point Q will ultimately approach the
fixed point P, until at length these points
may be made to coincide. When the straight
line PQ reaches this limiting position, it be-
comes the tangent to the circle at the point P.

Hence a tangent may be defined as a
straight line which passes through ttoo coin-
cident points on the circumference.

6. Circles are said to touch one another when they
meet, but do not cut one another.

Note. When each of the circles which meet is oiUside the other ^
they are said to touch one another externally, or to have external
contact : when one of the circles is within the other, the first is said
to touch the other internally, or to have internal contact with it.

7. A segment of a circle is the figure bounded by a
chord and one of the two arcs into which the chord divides
the circumference.

Note. The chord of a segment is sometimes called its base.

166 Euclid's elements.

8. An angle in a segment is one
formed by two straight lines drawn from
any point in the arc of the segment to
the extremities of its chord.

Note, (i) It will be shewn in Proposition 21, that all angles in
the same segment of a circle are equal.

Note, (ii) The angle o/a segment (as disunct from the angle in a
segment) is sometimes defined as that which is contained between
the cJiord and the arc ; but this definition is not required in any
proposition of Euclid.

9. An angle at the circumference of
a circle is one formed by straight lines
drawn from a point on the circumference
to the extremities of an arc : such an
angle is said to stand upon the arc by
which it is subtended.

10. Similar segments
of circles are those which
contain equal angles.

11. A sector of a circle is a figure
bounded by two radii and the arc inter-
cepted between them.

Symbols and Abbreviations.

In addition to the symbols and abbreviations given on
page 11, we shall use the following.

/(9r circle, 0**/(9r circumference.

book iii. prop. 1. 167

Proposition 1. Problem.
To find the centre of a given circle.

Let ABC be a given circle.
It is required to find the centre of the ABC.

Construction. In the given circle draw any chord AB,

and bisect AB at D. I. 10.

From D draw DC at right angles to AB ; I. 1 1.
and produce DC to meet the O ** at E and C.

Bisect EC at F. I. 10.

Then shall F be the centre of the O ABC.

Proof. First, the centre of the circle must be in EC :
for if not, suppose the centre to be at a point G outside EC.

Join AG, DG, BG.

Then in the A" GDA, GOB,

DA=DB, Constr,

and GD is common ;
and GA = GB, for by supposition they are radii;
.-. the iL GDA = the l GDB; I. 8.

. •. these angles, being adjacent, are rt. angles.

But the ^ CDB is a rt. angle ; Constr,

,\ the L GDB = the l CDB, Ax. 11.

the part equal to the whole, which is impossible.
.'. G is not the centre.
So it may be shewn that no point outside EC is the centre ;

.*. the centre lies in EC.

.'. F, the middle point of the diameter EC, must be the

centre of the ABC. Q.E.F.

Corollary. The straight line which bisects a chord of a
circle at right angles passes through the centre.

Because-

168 euclid's elements.

Proposition 2. Theorem.

If any two points are taken in the circvmference of a circle,
the dioid which joins them falls within the circle.

Let ABC be a circle, and A and B any two points on

its O'*.

Then shall the chord AB fall mthin the circle.

CJonstruction. Find D, the centre of the ABC ; IIL 1.
and in AB take any point E.
Join DA, DE, DB.

Proof. In the A DAB, because DA= DB, L Def 15.

.'. the L DAB = the L DBA. L 5.

But the ext. l DEB is greater than the int. opp. l DAE ;

L 16.
.'. the L DEB is also greater than the l DBE.

.'. in the A DEB, the side DB, which is opposite the greater
angle, is greater than DE which is opposite the less : I. 19.

that is to say, DE is less than DB, a radius of the circle ;

.*. E falls within the circle.

Similarly, any other point between A and B may be
shewn to fall within the circle.

.'. AB falls within the circle. q.e.d.

Note. A part of a curv^ed line is said to be concave to a point,
when for every chord (taken so as to lie between the point and the
curve) all straight lines drawn from the given point to the intercepted
arc are cut by the chord : if, when any chord whatever is taken, no
straight line drawn from the given point to the intercepted arc is
cut by the chord, the curve is said to be convex to that point.

Proposition 2 proves that the whole circumference of a circle is
concave to its centre.

book iii. prop. 3. 169

Proposition 3. Theorem.

If a straight line dratvn through the centre of a circle bisects
a chord which does not pass through the centre, it shall cut the
chord at right angles.

Conversely^ if it cuts the chord at right angles, it shall bisect it,

Let ABC be a circle; and let CD be a st. line drawn
through the centre, and AB a chord which does not pass
through the centre.

First, Let CD bisect the chord AB at F.
Then shall CD cut AB at rt, angles.

Construction. Find E the centre of the circle ; III. L

and join EA, EB.

Proot Then in the A" AFE, BFE,

[ AF = BF, Hyp,

Because i . and FE is common ;

[and AE = BE, being radii of the circle ;

.-. the 2. AFE = the z. BFE ; I. 8.

.*. these angles, being adjacent, are rt. angles;

that is, DC cuts AB at rt. angles. Q.E.D.

Conversely. Let CD cut the chord AB at rt. angles.
Then shall CD bisect AB at F.

Construction. Find E the centre ; and join EA, EB.

Proof. In the A EAB, because EA= EB, I. Def 15.

.-. the L EAB = the L EBA. I. 5.
Then in the A" EFA, EFB,

(the L EAF = the l EBF, Proved.

and the l EFA = the l EFB, being rt. angles; Eyp,
and EF is common ;

.-. AF=BF; I. 26.

that is, CD bisects AB at F. ^,^,\^.

170 EUCLID'S ELEMENTS.

Proposition 4. Theorem.

If in a circle two chords cut one another^ which do not both
pass through the centre, they cannot both be bisected at their point
of intersection.

Let ABCD be a circle, and AC, BD two chords which
intersect at E, but do not both pass through the centre.

Then AC and BD shall not be both bisected at E.

Case I. If one chord passes through the centre, it is a
diameter, and the centre is its middle point ;

.'.it cannot be bisected by the other chord, which by hypo-
thesis does not pass through the centre.

Case II. If neither chord passes through the centre ;
then, if possible, suppose E to be the middle point of both ;
that is, let AE = EC ; and BE = ED.

Construction. Find F, the centre of the circle. ill. 1,

Join EF.

Proof. Because FE, which passes through the centre,
bisects the chord AC, Hyp,

.*. the L FEC is a rt. angle. IIL 3.

And because FE, which passes through the centre, bi-
sects the chord BD, Hyp.

.*. the z. FED is a rt. angle. lii. 3,

.'. the L FEC == the l FED,

the whole equal to its part, which is impossible.

.•. AC and BD are not both bisected at E. Q.E.D.

EXERCISES ON PROPOSITIONS l-3o 171

EXERCISES.
ON Proposition 1.

1. If two circles intersect at the points A, B, shew that the line
which joins their centres bisects their common chord AB at right
angles.

2. AB, AC are two equal chords of a circle ; shew that the
straight line which bisects the angle BAC passes through the centre.

3. Two cho7'ds of a circle are given in position and magnitude :
find the centre of the circle,

4. Describe a circle that shall pass through three given points,
wMch are not in the same straight line.

5. Find the locus of the centres of circles which pass through two
given points.

6. Describe a circle that shall pass through two given points,
and have a given radius. When is this impossible ?

ON Proposition 2.
7* A straight line cannot cut a circle in inore than two points,

ON Proposition 3.

8. Through a given point within a circle draw a chord which
shall be bisected at that point.

9. The parts of a straight line intercepted between the circum-
ferences of two concentric circles are equal.

10. The line joining the middle points of two parallel chords of
a circle passes through the centre.

11. Find the locus of the middle points of a system of parallel
chords drawn in a circle.

12. If two circles cut one another, any two parallel straight lines
drawn through the points of intersection to cut the circles, are equal.

13. PQ and XY are two parallel chords in a circle : shew that
the points of intersection of PX, QY, and of PY, QX, lie on the
straight line which passes through the middle points of the given
chords.

172 EUCLID'S ELEMENTS.

Proposition 5. Theorem.

If two circles cut one another, they cannot have the same
centre.

Let the two O" AGO, BFC cut one another at C.
Then they shall not have the same centre.

Construction. If possible, let the two circles have the
same centre ; and let it be called E.

Join EC;
and from E draw any st. line to meet the O '^ at F and G.

Proof. Because E is the centre of the O AGC, Hyp.

.-. EG=EC.
And because E is also the centre of the O BFC, Hyp,

.-. EF=EC.

.-. EG=EF,

the whole equal to its part, which is impossible.

Therefore the two circles have not the same centre.

Q.E.D.

EXERCISES.
ON Propositions 4 and 5.

1. K a parallelogram can be inscribed in a circle, the point of
intersection of its diagonals must be at the centre of the circle.

2. Rectangles are the only parallelograms that can be inscribed
in a circle.

3. Two circles, which intersect at one point, must also intersect
at another.

BOOK III. PROP. 6. 178

Proposition 6. Theorem.

If two circles touch one another internally , they cannot have
the same centre.

Let the two 0* ABC, DEC touch one another internally
at C.

Then they shall not have the same centre.

Construction. If possible, let the two circles have the
same centre ; and let it be called F.

Join FC ;
and from F draw any st. line to meet the O"*" at E and B.

Proof. Because F is the centre of the O ABC, Hyp.

.-. FB = FC.

And because F is the centre of the O DEC, Hyp.

,', FE = FC.

.-. FB=FE,
the whole equal to its part, which is impossible.

Therefore the two circles have not the same centre.

Q.E.D.

Note. From Propositions 5 and 6 it is seen that circles, whose
circumferences have any point in common, cannot be concentric,
unless they coincide entirely.

Conversely, the circumferences of concentric circles can have no
point in common.

174 Euclid's elements.

Proposition 7. Theorem.

If from any 'point within a circle which is not the centre,
straight lines are drawn to the circumference, then the greatest
is that which passes through the centre ; and the least is the
remaining part of the diameter.

And of all other such lines, that which is nea/rer to the
greatest is always greater than one more remote.

And two equal straight lines, and only two, can he dravm
from the given point to the circumference, one on each side of the
diameter.

Let ABCD be a circle, and from F, any point within it
which is not the centre, let FA, FB, FC, FG, and FD be
drawn to the O **, of which FA passes through E the centre,
and FD is the remaining part of the diameter.

Then of all these st. lines,
(i) FA shall be the greatest ;
(ii) FD shall he the least ;
(iii) FB, which is nearer to FA, shall be greater than FC,

which is more remote;
(iv) also two, and only two, equal st, lines can be drawn from
F to the O **.'

Construction. Join EB, EC.

Proof, (i) In the A FEB, the two sides FE, EB are

together greater than the third side FB. i. 20.

But EB= EA, being radii of the circle ;

.'. FE, EA are together greater than FB ;

that is, FA is greater than FB.

BOOK III. PROP. 7. 175

Similarly FA may be shewn to be greater than any
other St. line drawn from F to the O ** ;

.'. FA is the greatest of all such lines.

(ii) In the A EFG, the two sides EF, FG are together

greater than EG ; i. 20.

and EG = ED, being radii of the circle ;

.'. EF, FG are together greater than ED.

Take away the common part EF ;

then FG is greater than FD.

Similarly any other st. line drawn from F to the O*^
may be shewn to be greater than FD ;

.'. FD is the least of all such lines.

(iii) In the A" BEF, CEF,

f BE = CE, I. i>6/. 15.

Because i and EF is common ;

[but the z_ BEF is greater than the z. CEF ;

.'. FB is greater than FG. I. 24.

Similarly it may be shewn that FG is greater than FG.

(iv) Join EG, and at E in FE make the l FEH equal to
the L FEG. I. 23.

Join FH.

Then in the A" GEF, HEF,

[ GE = HE, l,Def.l5.

Because - and EF is common ;

also the l GEF = the z. HEF; Constr.
/. FG = FH. I. 4.

And besides FH no other straight line can be drawn
from F to the C* equal to FG.

For, if possible, let FK = FG.

Then, because FH = FG, Proved.

.-. FK = FH,

that is, a line nearer to FA, the greatest, is equal to a line
which is more remote ; which is impossible. Proved,

Therefore two, and only two, equal st. lines can be
drawn from F to the O**. q.e.d.

176 euclid's elements.

Proposition 8. Theorem.

If from any point without a circle straight lines are dravm
to the circmnference^ of those which fall on the concave circum-
ference, the greatest is that which passes through the centre ; and
of others, that which is nearer to the greatest is always greater
than one more remote.

Of those which fall on the convex circmnference, the least is
that which, when produced, passes through the centre ; and of
others, that which is nearer to the least is always less than one
more remote.

From the given point there can he drawn to the drcamference
twOy and only two, equal straight lines, one on each side of the
shortest line.

Let BGD be a circle ; and from A, any point outside the
circle, let ABD, AEH, AFG, be drawn, of which AD passes
through C, the centre, and AH is nearer than AG to AD.
Then of st, lines drawn from A to the concave O *^,
(i) AD shall be the greatest, and (ii) AH greater than AG.
And of st, lines dravm from A to the convex O **,
(iii) AB shall be the least, and (iv) AE less than AF.
(v) Also two, and only two, equcd st, lines can be drawn
from A to the O*^.

Construction. Join CH, CG, CF, CE.
Proof, (i) In the A ACH, the two sides AC, CH are
together greater than AH : I. 20.

but CH = CD, being radii of the circle ;
.*. AC, CD are together greater than AH :
that is, AD is greater than AH.
Similarly AD may be shewn to be greater than any
other st. line drawn from A to the concave O "^ ;
.'. AD is the greatest of all such lines.

BOOK IH. PROP. 8. 177

(ii) In the A» HCA, GCA,

f HC = GC, I. Bef. 15.

Because \ and CA is common ;

[but the L HCA is greater than the l GCA ;

.'. AH is greater than AG. I. 24.

(iii) In the A AEC, the two sides AE, EC are together
greater than AC ; i. 20.

but EC=BC; L Def, 15.

.*. the remainder AE is greater than the remainder AB.

Similarly any other st. line drawn from A to the convex
O ^ may be shewn to be greater than AB ;

.'. AB is the least of all such lines.

(iv) In the A AFC, because AE, EC are drawn from the
extremities of the base to a point E within the triangle,

.*. AF, FC are together greater than AE, EC. I. 21.

But FC = EC; I. Def. 15.

.•. the remainder AF is greater than the remainder AE.

(v) At C, in AC, make the l ACM equal to the l ACE.

Join AM.

Then in the two A" ECA, MCA,

[ EC=MC, I. Def, 15.
Because - and CA is common ;

also the l ECA = the l MCA; Constr,

.-. AE = AM. I. 4.

And besides AM, no st. line can be drawn from A to the
O ", equal to AE.

For, if possible, let AK = AE :

then because AM = AE, Proved.

AM=AK;

that is, a line nearer to AB, the shortest line, is equal to
a line which is more remote ; which is impossible. Proved.

Therefore two, and only two, equal st. lines can be
drawn from A to the O**. Q.E.D.

Exercise. Where are the limits of that part of the circumfer-
ence which is concave to the point A ?
H.8.E. M

178 EUCLID'S ELEMENTS.

Proposition 9. Theorem. [First Proof.]

If from a point within a circle more than two equal straight
lines can he drawn to the drcmnference^ that point is the centre
of the circle.

Let ABC be a circle, and D a point within it, from which
more than two equal st. lines are drawn to the O*'*, namely
DA, DB, DC.

Then D shall he the centre of the circle ABC.

Construction. Join AB, BC :

and bisect AB, BC at E and F respectively. I. 10.

Join DE, DF.

Proot In the A' DEA, DEB,

(EA=EB, Constr,

and DE is common ;
and DA=DB; Hyp.

,\ the z. DEA = the Z- DEB; L 8.

»', these angles, being adjacent, are rt. angles.

Hence ED, which bisects the chord AB at rt. angles, must
pass through the centre. iii. 1. Cor,

Similarly it may be shewn that FD passes through the
centre.

.'. D, which is the only point common to ED and FD,
must be the centre. q.e.d.

Note. Of the two proofs of this proposition given by Euclid the
first has the advantage of being direct.

book iii. prop. 9. 179

Proposition 9. Theorem. [Second Proof.]

If from a point within a circle more than two equal straight
lines can he drawn to the circumference^ that point is the centre
of the circle.

Let ABC be a circle, and D a point within it, from which
more than two equal st. lines are drawn to the C^ namely
DA, DB, DC.

Then D sliall he the centre of the circle ABC.

Construction. For if not, suppose, if possible, E to be
the centre.

Join DE, and produce it to meet the O*'® at F, G.

Proof. Because D is a point within the circle, not the
centre, and because DF passes through the centre E ;

.*. DA, which is nearer to DF, is greater than DB, which
is more remote : ill. 7.

but this is impossible, since by hypothesis, DA, DB, are
equal.

.*. E is not the centre of the circle.

* And wherever we suppose the centre E to be, other-
wise than at D, two at least of the st. lines DA, DB, DC may
be shewn to be unequal, which is contrary to hypothesis.

.*. D is the centre of the ABC. Q.E.D.

*NoTE. For example, if the centre E were supposed to be
within the angle BDC, then DB would be greater than DA; if
within the angle ADB, then DB would be greater than DC ; if on
one of the three straight lines, as DB, then DB would be greater
than both DA and DC.

180 euclid's elements.

Proposition 10. Theorem. [First Proof.]
One circle cannot cut another at more than two points,

If possible, let DABC, EABC be two circles, cutting one
another at more than two points, namely at A, B, C.

Construction. Join AB, BC.

Draw FH, bisecting AB at rt. angles; L 10, 11.
and draw GH bisecting BC at rt. angles.

Proof. Because AB is a chord of both circles, and because
FH bisects AB at rt. angles,

.*. the centre of both circles lies in FH. iii. 1. Cor.

Again, because BC is a chord of both circles, and because
GH bisects BC at right angles,

.•. the centre of both circles lies in GH. III. 1. Cor.

Hence H, the only point common to FH and GH, is the
centre of both circles ;

which is impossible, for circles which cut one another
cannot have a common centre. ill. 5.

Therefore one circle caijnot cut another at more than two
points. Q.E.D.

Corollaries, (i) Two circles cannot have three points in
common without coinciding entirely.

(ii) Two circles cannot have a common arc without co-
inciding entirely.

(iii) Only one circle can be described through three points,
which are not in the same straight line.

BOOK III. PROP. 10. 181

Proposition 10. Theorem. [Second Proof.]
One circle cannot cut another at more than two points.

If possible, let DABC, EABC be two circles, cutting one
another at more than two points, namely at A, B, C.

Construction. Find H, the centre of the O DABC, III. 1.

and join HA, HB, HC.

Prool Since H is the centre of the O DABC,

.*. HA, HB, HC are all equal. I. Def. 15.

And because H is a point within the O EABC, from
which more than two equal st. lines, namely HA, HB, HC
are drawn to the O *^,

.-. H is the centre of the O EABC : III. 9.

that is to say, the two circles have a common centre H ;
but this is impossible, since they cut one another. III. 6.

Therefore one circle cannot cut another in more than
two points. Q.E.D.

Note. Both the proofs of Proposition 10 given by Euclid are
indirect.

The second of these is imperfect, because it assumes that the
centre of the circle DABC must fall within the circle EABC ;
whereas it may be conceived to fall either without the circle EABC,
or on its circumference. Hence to make the proof complete, two
additional cases are required.

182 euclid's elements.

Proposition 11. Theorem.

// two circles touch one another internally^ the straight line
which joins their centres, being produced, shail pass through the
point of contact.

Let ABC and ADE be two circles which touch one
another internally at A ; let F be the centre of the ABC,
and G the centre of the ADE.

Then shall FG produced pass through A.

Construction. For if not, suppose, if possible, FG to pass
otherwise, as FGEH.

Join FA, GA.

Proof In the A FGA, the two sides FG, GA are together
greater than FA : i. 20.

but FA= FH, being radii of the ABC : Ht/p,

FG, GA are together greater than FH.
Take away the common part FG :
then GA is greater than GH.

But GA = GE, being radii of the ADE : Hyp.

.'. GE is greater than GH,
the part greater than the whole ; which is impossible.
.*. FG, when produced, must pass through A.

Q.E.D.

exercises.

1. If the distance between the centres of two circles is equal to
the difference of their radii, then the circles must meet in one point,
but in no other ; that is, they must touch one another.

2. If two circles whose centres are A and B touch one another
internally J and a straight line is drawn through their point of contact^
cutting the circumferences at P and Q; shew that the radii AP and BQ
are parallel.

book iii. prop. 12. 183

Proposition 12. Theorem.

If two circles touch one another externally^ the straight line
which joins their centres shall pass through the point of contact.

Let ABC and ADE be two circles which touch one
another externally at A ; let F be the centre of the ABC,
and G the centre of the O ADE.

Then shall FG pass through A.

Construction. For if not, suppose, if possible, FG to pass
otherwise, as FHKG.

Join FA, GA.

Proof. In the A FAG, the two sides FA, GA are together
greater than FG : I. 20.

but FA= FH, being radii of the ABC ; Hyp.

and GA = GK, being radii of the ADE ; Hyp.

.'. FH and GK are together greater than FG;
which is impossible.
.'. FG must pass through A.

Q.E.D.

EXERCISES.

1. Find the locus of the centres of all circles which touch a given
circle at a given point.

2. Find the locus of the centres of all circles of given radius^ which
touch a given circle.

3. If the distance between the centres of two circles is equal to
the sum of their radii, then the circles meet in one point, but in no
other ; that is, they touch one another.

4. If two circles whose centres are A and B touch one another
externally y and a straight line is drawn through their point of contact
cutting the circumferences at P and Q; shew that the radii AP and BQ
are XKuraUd.

184

Euclid's elements.

Proposition 13. Theorem.

Two circles cannot touch one another at more than one point,
whether internally or externxdly.

Fig. 2.

If possible, let ABC, EDF be two circles which touch one
another at more than one point, namely at B and D.

Construction. Join BD ;

and draw GF, bisecting BD at rt. angles. I. 10, 11.

Proof. Now, whether the circles touch one another
internally, as in Fig 1 or externally as in Fig 2,

because B and D are on the O "^ of both circles,

.-. BD is a chord of both circles :

.*. the centres of both circles lie in GF, which bisects
BD at rt. angles. III. 1. Cor.

Hence GF which joins the centres must pass through
a point of contact ; III. 11, and 12.

which is impossible, since B and D are outside GF.

Therefore two circles cannot touch one another at more
than one point.

Q.E.D.

Note. It must be observed that the proof here given applies,
by virtue of Propositions 11 and 12, to both the above figures : we
have therefore omitted the separate discussion of Fig. 2, which finds
a place in most editions based on Simson's text.

BOOK III. PROP. 13. 185

EXERCISES ON PROPOSITIONS 1-13.

1. Describe a circle to pass through two given points and have
its centre on a given straight line. When is this impossible ?

2. All circles which pass through a fixed point, and have their
centres on a given straight line, pass also through a second fixed
point.

3. Describe a circle of ^iven radius to touch a given circle at a
given point. How many solutions will there be ? When will there
be only one solution ?

4. From a given point as centre describe a circle to touch a
given circle. How many solutions will there be ?

5. Describe a circle to pass through a given point, and touch a
given circle at a given point. [See Ex. 1, p. 183, and Ex. 5, p. 171.]
When is this impossible ?

6. Describe a circle of given radius to touch two given circles.
[See Ex. 2, p. 183.] How many solutions will there be?

7. Two parallel chords of a circle are six inches and eight inches
in length respectively, and the perpendicular distance between them
ifi one inch : find the radius.

8. If two circles touch one another externally, the straight lines^
which join the extremities of parallel diameters towards opposite
parts, must pass through the point of contact.

9. Find the greatest and least straight lines which have one
extremity on each of two given circles, which do not intersect.

10. In any segment of a circle, of all straight lines drawn at
right angles to the chord and intercepted between the chord and the
arc, the greatest is that which passes through the middle point of
the chord ; and of others that which is nearer the greatest is greater
than one more remote.

11. If from any point on the circumference of a circle straight
lines be drawn to the circumference, the greatest is that which passes
through the centre ; and of others, that which is nearer to the greatest
is greater than one more remote ; and from this point there can be
drawn to the circumference two, and only two, equal straight lines.

186

EUCLID'S ELEMENTS.

Proposition 14. Theorem.

Equal ch&rds in a circle are equidistant from the centre.
Conversely, chords which are equidistant from the centre
are equal.

Let ABC be a circle, and let AB and CD be chords, of
which the perp. distances from the centre are EF and EG.

First. LetAB = CD.

Then shall AB and CD be equidistant from the centre E.

Construction. Join EA, EC.

Proof. Because EF, which passes through the centre,

Hyp,
IIL 3.

Hyp,
Ax, 7.

is perp. to the chord AB ;

.*. EF bisects AB;
.*. AB is double of FA.

For a similar reason, CD is double of GC.
But AB = CD,
.-. FA = GC.

Now EA = EC, being radii of the circle ;
.'. the sq. on EA = the sq. on EC.

But since the l E FA is a rt. angle :
.'. the sq. on EA = the sqq. on EF, FA.

And since the l EGC is a rt. angle ;
.'. the sq. on EC = the sqq. on EG, GC.
.-. the sqq. on EF, FA = the sqq. on EG, GC.

Now of these, the sq. on FA = the sq. on GC ; for FA = GC.
.•. the sq. on EF = the sq. on EG;
.-. EF=EG;

that is, the chords AB, CD are equidistant from the centre.

Q.E.D.

L 47.

BOOK III. PROP. 14. 187

Conversely, Let AB and CD be equidistant from the
centre E ;

that is, let EF=EG.
Then shall AB = CD.

Proof. The same construction being made, it may be
shewn as before that AB is double of FA, and CD double
of GC ;

and that the sqq. on EF, FA = the sqq. on EG, GC.
Now of these, the sq. on EF = the sq. on EG,

forEF=EG: Hyp,

.-. the sq. on FA = the sq. on GC;
.-. FA = GC;
and doubles of these equals are equal ; Ax, 6.
that is, AB = CD.

Q.E.D.

EXERCISES.

1. Find the locus of the middle points of equal chords of a circle,

2. If two chords of a circle cut one another, and make equal
angles with the straight line which joins their point of intersection
to the centre, they are equal.

3. If two equal chords of a circle intersect, shew that the segments
of the one are eqvxil respectively to the segments of the other.

4. In a given circle draw a chord which shall be equal to one
given straight line (not greater than the diameter) and parallel to
another.

5. PQ is a fixed chord in a circle, and AB is any diameter :
shew that the sum or diflference of the perpendiculars let fall from
A and B on PQ is constant, that is, the same for all positions of AB.

188 euclid's elements.

Proposition 15. Theorem.

The diameter is the greatest chord in a circle ;

and of other chwds, that which is nearer to the centre is
greater than one more remote.

Conversely^ the greater chord is nearer to the centre than
the less.

Let ABCD be a circle of which AD is a diameter, and E
the centre ; and let BC and FG be any two chords, whose
perp. distances from the centre are EH and EK.

Then (i) AD shall be greater than BC ;

(ii) if EH is less than EK, BC shall he greater than FG :
(iii) if BC is greater than FG, EH shall be less than EK.

(i) Construction. Join EB, EC.

Proof. In the A BEC, the two sides BE, EC are
together greater than BC ; i. 20.

but BE = AE, I. Def 15.

and EC=ED;
.'. AE and ED together are greater than BC ;
that is, AD is greater than BC.

Similarly AD may be shewn to be greater than any
other chord, not a diameter.

(ii) Let EH be less than EK.

Then BC shall be greater than FG.

Construction. Join EF.

Proof. Since EH, passing through the centre, is perp.
to the chord BC,

.-. EH bisects BC ; m. 3.

BOOK in. PROP. 15. 189

.-. BC is double of HB.
For a similar reason FG is double of KF.

NowEB=EF, I. J9e/. 15.

.*. the sq. on EB = the sq. on EF.

But since the z_ EHB is a rt. angle,
.-. the sq. on EB = the sqq. on EH, HB. I. 47.

And since the z_ EKF is a rt. angle,

.'. the sq. on EF = the sqq. on EK, KF;

.'. the sqq. on EH, HB = the sqq. on EK, KF.

But the sq. on EH is less than the sq. on EK,

for EH is less than EK ; Hyp,

,\ the sq. on HB is greater than the sq. on KF ;
.-. HB is greater than KF :
hence BC is greater than FG.

(iii) Canmrsely. Let BC be greater than FG.

Then EH shall he less than EK.

Prool The same construction being made, it may be
shewn as before that BC is double of BH. and FG double of
FK; and that the sqq. on EH, HB = the sqq. on EK, KF.

But since BC is greater than FG, Hyp.

,\ HB is greater than KF ;
••. the sq. on HB is greater than the sq on KF.
.*. the sq. on EH is less than the sq. on EK ;

.'. EH is less than EK. Q.E.D.

EXERCISES.

1. Through a given point within a circle draw the least possible
chord.

2. AB is a fixed chord of a circle, and XY any other chord
having its middle point Z on AB ; what is the greatest, and what
the least length that X Y may have ?

Shew that XY increases, as Z approaches the middle point of AB.

3. In a given circle draw a chord of given length, having its
middle point on a given chord.

When is this pr^lem impossible ?

190

EUCLID'S ELEMENTS.

Proposition 16. Theorem. [Alternative Proof.]

The straight line drawn at right angles to a diameter of
a circle at one of its extremities is a tangent to the circle :

and every other straight line drawn through this point cuts
the circle,

C G

Let AKB be a circle, of which E is the centre, and AB
a diameter ; and through B let the st. line CBD be drawn
at rt. angles to AB.

Then (i) CBD shall he a tangent to the circle ;

(ii) any other st, line through B, such as BF, shall cut
the circle,

(i) Construction. In CD take any point G, and join EG.

Proof. In the A EBG, the L EBG is a rt. angle ; Hyp.
,\ the L EGB is less than a rt. angle ; l. 17.

.'. the Z- EBG is greater than the z. EGB ;

.*. EG is greater than EB : L 19.

that is, EG is greater than a radius of the circle ;
.*. the point G is without the circle.
Similarly any other point in CD, except B, may be
shewn to be outside the circle.

Hence CD meets the circle at B, but being produced,
does not cut it ;

that is, CD is a tangent to the circle, ill. Def 5.

(ii) Construction. Draw EH perp. to BF. L 12.

Proof. In the A EHB, because the z_ EHB is a rt. angle,

.*. the L EBH is less than a rt. angle ; i. 17.

.-. EB is greater than EH ; L 19.

BOOK III. PROP. 16. 191

that is, EH is less than a radius of the circle :
.*. H, a point in BF, is within the circle ;

.*. BF must cut the circle. Q.E.D.

Note. The above proof of Proposition 16 is not that given by
Euclid, but it is preferable as being direct. Euclid's proof by
Beductio ad Ahsurdum is given below.

Proposition 16. Theorem. [Euclid's Proof.]

The straight line dratvn at right angles to a diameter of a circle a^
one of its extremities, is a tafigent to the circle :

and no other straight line can be dravm through this point so cw
not to cut the circle.

Let ABC be a circle, of which D is the centre, and AB a diameter ;
let AE be drawn at rt. angles to BA, at its extremity A.

(1) TIten shall AE &e a tangent to the circle.

Construction.

For, if possible, suppose AE to cut the circle at C.

Join DC.

Proof. Then in the A DAC, because DA= DC, i. Def, 15.

.-. the L DAC = the L DCA.

But the L DAC is a rt. angle ; ffyp-

.*. the L DCA is a rt. angle ;

that is, two angles of the A DAC are together equal to two rt.
angles ; which is impossible. i. 17.

Hence AE meets the circle at A, but being produced, does not
cut it ;

that is, AE is a tangent to the circle. iii. Def, 6.

192 EUCLID'S ELEMENTS.

(ii) Also through A no other straight line but AE can be drawn so
as not to cut tJie circle.

Construction. For, if possible, let A F be another st. line drawn
through A so as not to cut the circle.

From D draw DG perp. to AF ; i. 12.

and let DG meet the O*^ at H.

Proof. Then in the A DAG, because the L DGA is a rt. angle,
.'. the L DAG is less than a rt. angle ; i. 17.

.*. DA is greater than DG. i. 19.

But DA=DH, I. Def, 15.

.*. DH is greater than DG,
the part greater than the whole, which is impossible.

Therefore no st. line can be drawn from the point A, so as not to
cut the circle, except AE.

Corollary, (i) A tangent touches a circle at one point only.

Corollary, (ii) Tliere can be but one tangent to a circle at a
given point.

book iii. prop. 17. 193

Proposition 17. Problem.

To draw a tangent to a circle from a given 'point either' on, or
laithout the circumference.

Fig. X.

'Oy

Let BCD be the given circle, and A the given point.
It is required to draw from A a tangent to the G CDS,

Cask I. When the given point A is on the O**.

Construction. Find E, the centre of the circle. ill. 1.

Join EA
At A draw AK at rt. angles to EA. I. 11.

Proof. Then AK being perp to a diameter at one of its
extremities, is a tangent to the circle. iii. 16.

Case II. When the given point A is without the O**.

Construction. Find E, the centre of the circle; ill. 1.

and join AE, cutting the O BCD at D.

With centre E and radius EA, describe the O AFG.

At D, draw GDF at rt. angles to EA, cutting the O AFG

at F and G. I. 11.

Join EF, EG, cutting the BCD at B and C.

Join AB, AC.
Then both AB and AC shall be tangents to the CDB.

Proot In the A'AEB, FED,

(AE = FE, being radii of the QAF ;
and EB= ED, being radii of the BDC ;
and the included angle AEF is common ;

.-. the ^ ABE == the z. FDE. 1.4.

H.S.B. Xi

194

EUCLID S ELEMENTS.

But the z_ FDE is a rt. angle ; Constr.

.'. the L ABE is a rt. angle.

Hence AB, being drawn at rt. angles to a diameter at one
of its extremities, is a tangent to the O BCD. ill. 16.

Similarly it may be shewn that AC is a tangent. Q.E.F.

Corollary. If two tangents are drawn to a circle from an
external j^ointj then (i) they are equal; (ii) they mibtend equal angles
al the centre ; (iii) they make equal angles with the straight line which
joins the given point to the centre.

For, in the above figure,

Since ED is perp. to FQ, a chord of the FAG, Gonstr.

:. DF = DG. m. 3.

Then in the A« DEF, DEG,

TDE is common to both,

Because-! and EF = EG; i. Def 15.

\ andDF = DG; Proved.

.'. the L DEF = the L DEQ. l 8.

Again in the A* AEB, AEC,
AE is common to both,
Because - and EB = EC,

and the L AEB = the L AEC ; Proved.

.'. AB=AC: I. 4.

and the L EAB=the L EAC. q.e.d.

Note. If the given point A is within the circle, no solution is
possible. Hence we see that this problem admits of two solutions,
one solution, or no solution, according as the given point A is without,
on, or within the circumference of a circle. For a simpler method
of drawing a tangent to a circle from a given point, see page 218.

book iii. prop. 18. 195

Proposition 18. Theorem.

The straight line dravm from the centre of a circle to the
point of contact of a tangent is perpendicular to the tangent.

Let ABC be a circle, of which F is the centre ;
and let the st. line DE touch the circle at C.

Then shall FC be perp, to DE.

For, if not, suppose, if possible, FQ to be perp. to DE, 1. 12.

and let FG meet the O** at B.
Proof.
In the A FCG, because the l FGC is a rt. angle. Hyp,
.*. the L FCG is less than a rt. angle ; I. 17.

.'. the L FGC is greater than the l FCG ;

.'. FC is greater than FG : I. 19.

but FC = FB ;

.*. FB is greater than FG,

the part greater than the whole, which is impossible.

.*. FC cannot be otherwise than perp. to DE :

that is, FC is perp. to DE. Q.E.D.

EXERCISES.

1. Draw a tangent to a circle (i) parallel to, (ii) at right angles
to a given straight line.

2. Tangents draion to a circle from the extremities of a diameter
are paraMel,

3. Circles which touch one another internally or externally have a
common tangent at their point of contact.

4. In two concentric circles ^ any chord of the outer circle which
touches tJte inner j is bisected at the point of contact.

5. In two concentric circles, all chords of the outer circle which
touch the inner, are equal.

196

euclid's elements.
Proposition 19. Theorem.

The straight line drawn perpendicular to a tangent to a
circle from the point of contact passes through the centre.

Let ABC be a circle, and DE a tangent to it at the point
C ; and let CA be drawn perp. to DE.

Then shall CA pass through the centre.

Construction. For if not, suppose, if possible, the centre
F to be outside CA.

Join CF.

Proof. Because DE is a tangent to the circle, and FC
is drawn from the centre F to the point of contact,

.*. the L FCE is a rt. angle. in. 18.

But the z. ACE is a rt. angle ; Hyp,

.*. the L FCE = the l ACE ;
the part equal to the whole, which is impossible.

.*. the centre cannot be otherwise than in CA ;
that is, CA passes through the centre.

Q.KD.

exercises on the tangent.
Propositions 16, 17, 18, 19.

1. The centre of any circle which touches two intersecting straight
lines must lie on the bisector of the angle between them,

2. AB and AC are two tangents to a circle whose centre is O;
shew that AO bisects the chord of contact BC at right angles.

EXERCISES ON THE TANGENT. 197

3. If two circles are concentric all tangents drawn from points
on the circumference of the outer to the inner circle are equal.

4. The diameter of a circle bisects all chords which are parallel
to the tangent at either extremity.

5. Find the locus of the centres of all circles which touch a given
straight line ai a given point.

6. Find the locus of the centres of all circles which touch each
of two parallel straight lines.

7. Find the loctis of the centres of all circles which touch each of
two intersecting straight lines of unlimited length,

8. Describe a circle of given radius to touch two given straight
lines.

9. Through a given point, within or without a circle, draw a
chord equal to a given straight line.

In order that the problem may be possible, between what limits
must the given line lie, when the given point is (i) without the circle,
(ii) within it ?

10. Two parallel tangents to a circle intercept on any third
tangent a segment which subtends a right angle at the centre.

11. In any quadrilateral circumscribed about a circle^ the sum of
one pair of opposite sides is equal to the sum of the other pair,

12. Any parallelogram which can be circumscribed about a
circle, must be equilateral.

13. If a quadrilateral is described about a circle, the angles
subtended at the centre by any two opposite sides are together equal
to two right angles.

14. AB is any chord of a circle, AC the diameter through A,
and AD the perpendicular on the tangent at B : shew that AB
bisects the angle UAC.

15. Find the locus of the extremities of tangents of fixed length
drawn to a given circle.

16. In the diameter of a circle produced, determine a point
such that the tangent drawn from it shall be of given length.

17. In the diameter of a circle produced, determine a point
such that the two tangents drawn from it may contain a given angle.

18. Describe a circle that shall pass through a given point, and
touch a given straight line at a given point. [See page 197. Ex. 5.]

19. Describe a circle of given radius, having its centre on a
given straight line, and touching another given straight line.

20. Describe a circle that shall have a given radius, and touch a
given circle and a given straight line. How many such circles can
be drawn ?

198 euclid's elements.

Proposition 20. Theorem.

The angle at the centre of a circle is double of an angle at
the circumference, standing on the same arc.

Fig. I. . Fig. 2.

Let ABC be a circle, of which E is the centre ; and let
BEC be the angle at the centre, and BAG an angle at the
O**, standing on the same arc BC.

Then shall the l BEC he dovhle of the l BAC.
Construction. Join AE, and produce it to F.

Case I. When the centre E is within the angle BAC.

Proof. In the A EAB, because EA = EB, I. Def 15.

.-. the ^ EAB = the L EBA; L 5.
.*. the sum of the l* EAB, EBA = twice the l EAB.

But the ext. l BEF = the sum of the z_" EAB, EBA; L 32.

.*. the L BEF = twice the l EAB.
Similarly the l FEC = twice the l EAC.
.-. the sum of the z.' BEF, FEC = twice the sum of

the ^" EAB, EAC ;
that is, the £. BEC = twice the l BAC.

Case II. When the centre E is without the l BAC.

As before, it may be shewn that the z. FEB = twice the l FAB;

also the z. FEC = twice the l FAC;
.*. the difference of the z_' FEC, FEB = twice the difference
of the ^' FAC, FAB :

that is, the l BEC = twice the l BAC. Q.E.D.

BOOK III. PROP. 21.

Note 1. The case in which the centre E
falk 071 AB or AC needs no proof beyond that
given under Case I.

Note 2. If the arc BFC, on which the
angles stand, is greater than a semi- circum-
ference, the angle BEC at the centre will be
reflex: but it may still be shewn as, in Case I.,
that the reflex z BEC is double of the z BAG
at the O^, standing on the same arc BFC.

199

Proposition 21. Theorem.
Angles in the same segment of a circle are equal.

Let ABCD be a circle, and let BAD, BED be angles in
the same segment BAED.

Then shall the l BAD = the l BED.

Construction. Find F, the centre of the circle. III. 1.

Case I. When the segment BAED is greater than a
semicircle.

Join BF, DF.

Proof. Because the l BFD is at the centre, and the
L. BAD at the O**, standing on the same arc BD,

.'. the L BFD = twice the l BAD. ill. 20.

Similarly the l BFD = twice the /_ BED. III. 20.

.-. the ^ BAD = the iL BED. Ax. 7,

200 fitJCLID^S ELEMENTS.

Case II. When the segment BAED is not greater than
a semicircle.

Construction.

Join AF, and produce it to meet the O"* at C.

Join EC.

Proof. Then since AEDC is a semicircle ;
.*. the segment BAEC is greater than a semicircle :
.*. the L BAC = the l BEC, in this segment. Case. 1

Similarly the segment CAED is greater than a semicircle ;
.*. the L CAD = the z- CED, in this segment.

.*. the iJ BAC, CAD = the sum of the iJ BEC, CED .
that is, the l BAD = the l. BED. Q.E.D.

EXERCISES.

1. P is any point on the arc of a segment of which AB is the
chord. Shew that the sum of the angles PAB, PBA is constant.

2. PQ and RS are two chords of a circle intersecting at X :
prove that the triangles PXS, RXQ are equiangular.

3. Two circles intersect at A and B ; and through A any straight
line PAQ is drawn terminated by the circumferences : shew that PQ
subtends a constant angle at B.

4. Two circles intersect at A and B ; and through A any two
straight lines PAQ, XAY are drawn terminated by the circum-
ferences ; shew that the arcs PX, QY subtend equal angles at B.

5. P is any point on the arc of a segment whose chord is AB :
and the angles PAB, PBA are bisected by straight lines which
intersect at O. Find the locus of the point O.

BOOK III. PROP. 21.

201

Note. If the extension of Proposition 20, given in Note 2 on
page 199, is adopted, a separate treatment of
the second case of the present proposition is —

unnecessary.

For, as in Case I. ,

the reflex L BFD= twice the L BAD ;

III. 20.

also the reflex L BFD = twice the L BED ;

.*. the L BAD = the L BED.

Ohs, The converse of Prop. 21 is important. For the construc-
tion used, viz. To describe a circle ahout a given triangle, se.e Book
rv., Prop. 5, or Theorems and Examples on Book i, page 111, No. 1.

Converse of Proposition 21.

Equal angles standing on the same base, and on the same side of it,
have their vertices on an arc of a circle, of which the given base is the
chord.

Let BAC, BDC be two equal angles standing
on the same base BC.

Then shaU the vertices A and D lie upon a
segment of a circle having BC as its chord.

Describe a circle about the A BAC. iv. 6.

Then this circle shall pass through D.

For, if not, it must cut BD, or BD produced,
at some other point E.

Join EC.

Then the L BAC = the L BEC, in the same segment : iii. 21.
but the L BAC = the L BDC, by hypothesis ;
.-. the L BEC = the L BDC ;
that is, an ext. angle of a triangle = an int. opp. angle ;

which is impossible. i. 16.

.*. the circle which passes through B, A, C, cannot pass other-
wise than through D.

That is, the vertices A and D are on an arc of a circle of which
the chord is BC. q.e.d.

Corollary. The locus of the vertices of triangles drawn on the
same base and on the same side of it with equal vertical angles is an
arc of a circle.

202 euclid's elements.

Proposition 22. Theorem.

The opposite angles of any quadrilateral inscribed in a circle
are together equal to two right angles.

Let ABCD be a quadrilateral inscribed in the O ABC.

Then shall (i) the iJ ADC, ABC together — two rt. angles ;
(ii) the L} BAD, BCD together = two rt, angles.

Construction. Join AC, BD.

Proof.
Since the l ADB = the l ACB, in the segment ADCB; IIL 21,
and the l CDB = the l CAB, in the segment CDAB;
.-. the L ADC = the sum of the ^" ACB, CAB.

To each of these equals add the l ABC :
then the two l* ADC, ABC together = the three l" ACB,
CAB, ABC.

But the L" ACB, CAB, ABC, being the angles of a triangle,
together = two rt. angles ; i. 32.

.'. the L' ADC, ABC together = two rt. angles.
Similarly it may be shewn that

the jl" bad, BCD together = two rt. angles. Q.E.D.

exercises.

1. If a circle can be described about a parallelogram, the paral-
lelogram must be rectangular.

2. ABC is an isosceles triangle, and XY is drawn parallel to the
base BC cutting the sides in X and Y : shew that the four points B,
C, X, Y lie on a circle.

3. If one side of a quadrilateral inscribed in a circle is producedf
the exterior angle is equal to the opposite interior angle of the quadri-
lateral.

BOOK III. PROP. 22.

203

Proposition 22. [Alternative Proof.]

Let ABCD be a quadrilateral inscribed in the O ABC.

Then shall the Z." ADC, ABC together = two rt. angles.

Join FA, FC.

Since the L AFC at the centre = twice the

L. ADC at the O*®, standing on the same arc

ABC ; III. 20.

and the reflex angle AFC at the centre

= twice the L ABC at the O**, standing on the

same arc ADC ; in. 20.

.-. the L* ADC, ABC are together half

the sum of the Z. AFC and the reflex angle AFC;

but these make up four rt. angles : i. 15. Cor. 2.
.*. the L* ADC, ABC together = two rt. angles, q.e.d.

Definition. Four or more points through which a
circle may be described are said to be concyclic.

Converse of Proposition 22.

If a pair of opposite angles of a quadrilateral are together equal
to two right angles^ its vertices are concyclic.

Let ABCD be a quadrilateral, in which the opposite angles at
B and D together = two rt. angles.

Then shall the four points A, B, C, D he
concyclic.

Through the three points A, B, C describe
a circle. iv. 5.

Then this circle must pa^ss through D.
For, if not, it will cut AD, or AD produced,
at some other point E.

Join EC.
Then since the quadrilateral ABCE is inscribed in a circle,

/. the Z." ABC, AEC together = two rt. angles. iii. 22.

But the Z." ABC, ADC together = two rt. angles ; ffyp-

hence the Z." ABC, AEC = the Z."ABC, ADC.

Take from these equals the L ABC ;

then the L AEC=the L ADC ;

that is, an ext. angle of a triangle = an int. opp. angle ;

which is impossible. i. 16.

.*. the circle which passes through A, B, C cannot pass otherwise
than through D :

that is the four vertices A, B, C, D are concyclic. Q.E.D.

204 EUCLID'S ELEMENTS.

Definition. Similar segments of circles are those
which contain equal angles. [Book ill., Def. 10.]

Proposition 23. Theorem.

On the same chord and on the same side of itj there cannot
he two similar segments of circles^ not coinciding with one
another.

If possible, on the same chord AB, and on the same
side of it, let there be two similar segments of circles ACB,
ADB, not coinciding with one another.

Then since the arcs ADB, ACB intersect at A and B,

.*. they cannot cut one another again; ill. 10.
.*. one segment falls within the other.

Construction. In the inner arc take any point 0.
Join AC, producing it to meet the outer arc at D :

join CB, DB.

Proof. Then because the segments are similar,

.-. the L ACB = the l ADB; in. Def 10.

that is, an ext. angle of the A CDB = an int. opp. angle;

which is impossible. i. 16.

Hence the two similar segments ACB, ADB, on the same
chord AB and on the same side of it, must coincide. Q.E.D.

EXERCISES ON PROPOSITION 22.

1. The straight lines which bisect any angle of a quadrilateral
figure inscribed in a circle and the opposite exterior angle, meet on
the circumference.

2. A triangle is inscribed in a circle : shew that the sum of the
angles in the three segments exterior to the triangle is equal to four
right angles.

.3. Divide a circle into two segments, so that the angle contained
by the one shall be double of the angle contained by the other.

BOOK III. PROP. 24. 205

Proposition 24. Theorem.

Similar segments of circles on equal chords are equal to one
another,

£ ^ O^

Let AEB and CFD be similar segments on equal chords
AB, CD.

Then shall the segment AEB = the segment CFD.

Proof. If the segment AEB be applied to the segment
CFD, so that A falls on C, and AB falls along CD ;

then since AB = CD, Hyp.

.*. B must coincide with D.

.'. the segment AEB must coincide with the segment
CFD ; for if not, on the same chord and on the same side of
it there would be two similar segments of circles, not co-
inciding with one another : which is impossible. ill. 23.

.*. the segment AEB = the segment CFD. Q.E.D.

EXERCISES.

1. Of two segments standing on the same chord, the greater
segment contains the smaller angle.

2. A segment of a circle stands on a chord AB, and P is any
point on the same side of AB as the segment : shew that the angle
APB is greater or less than the angle in the segment, according as P
is within or without the segment.

3. P, Q, R are the mid^He points of the aides of a triangle, and X
is the foot of the perpendictUar let fall from one vertex on the opposite
side : shew thai the four points P, Q, R, X are coney die.

[See page 104, Ex. 2: also page 108, Ex. 2.]

4. Use the preceding exercise to shew that the middle points of the
sides of a triangle and the feet of the perpendiculars let fall from the
vertices on the opposite sides, are concyclic.

206 EUCLID'S ELEMENTS.

Proposition 25. Problem.

An arc of a circle being given^ to describe the whole circum-
ference of which the given arc is a part.

Let ABC be an arc of a circle.

It is required to describe the whole O®* of which the arc ABC is a
part.

Construction.

In the given arc take any three points A, B, C.

Join AB, BC.
Draw DF bisecting AB at rt. angles, l. 10. 11.
and draw EF bisecting BC at rt. angles.

Proof.
Then because DF bisects the chord AB at rt. angles,

.'. the centre of the circle lies in DF. III. 1 Cor.
Again, because EF bisects the chord BC at rt. angles,

.*. the centre of the circle lies in EF. iii. 1 Cor.

.'. the centre of the circle is F, the only point common to
DF and EF.

Hence the O** of a circle described with centre F, and
radius FA, is that of which the given arc is a part. Q.E.F.

BOOK III. PROP. 25.

207

Note. Euclid gave to this proposition a somewhat different
form, as follows :

Proposition 25. [Euclid's Method.]

A segment of a circle being given, to describe the circle of which it is
a segment.

Let ABC be the given segment of a circle, standing on the chord
AC.

It is required to describe the circle of which ABC is a segment.

Construction. Draw DB, bisecting AC at rt. angles, and meeting
the 0« at B.

Join AB.

Case I. When the L DAB is not equal to the L ABD.
At A, in BA, make the L BAE equal to the L ABD ; I. 23.

and let AE meet BD, or BD produced, at E.

Join EC.

Tlien E sh^all be the centre of the required circle.

Proof. Since the L EAB = the L EBA, Gonstr.

:. EA=EB. I. 6.
And in the A» EDA, EDC,

{DA=DC, Constr.
and ED is common ;
also the L EDA=the L EDC, being rt. angles ;

.-. EA=EC. I. 4.

Hence EA, EB, and EC are all equal ;
.•. E is the centre of the required circle, and EA, EB, EC are radii.

Case II. When the L DAB = the L ABD.

In this case it follows that DB = DA ; I. 6.
.*. DB, DA, DC are all equal, so that D is the
centre of the required circle.

208 Euclid's elements.

Proposition 26. Theorem.

In equal circles the arcs which subtend equal angles, whether
at the centres or at the circumferences, shall he egual.

Let ABC, DEF be equal circles ;

and let the z_* BGC, EHF at the centres be equal,

and consequently the z." BAG, EDF at the O*"^ equal. IIL 20.

Then shall the arc BKG = the arc ELF.

Construction. Join BC, EF.

Proof. Because the O* ABC, DEF are equal,

.". their radii are equal.

Hence in the A' BGC, EHF,
[ BG = EH,

Because-! and GC = HF,

[and the l BGC = the z_ EHF ; Hyp,

.-. BC = EF. L 4.

Again, because the l BAC = the l EDF, Hyp»

.*. the segment BAG is similar to the segment EDF ;

III. Def. 10.
and these segments are on equal chords BC, EF ;
.'. the segment BAG = the segment EDF. III. 24.

But the whole ABC = the whole DEF;
.•. the remaining segment BKC = the remaining segment ELF;

.-. the arc BKC = the arc ELF.

Q.E.D.

[For Exercises and an Alternative Proof see pp. 212, 213.]

BOOK III. PROP. 27. 209

Proposition 27. Theorem.

In equal circles the angles, ichether at the centres or the cir-
cumferenceSy which stand on equal arcs, shall be equal.

Let ABC, DEF be equal circles ;
and let the arc BC = the arc EF.

Then shall the l BGC = the l EHF, at the centres ;
and also the l BAG = the l EDF, at the O^,

Construction. If the iJ BGC, EHF are not equal, one
must be the greater.

If possible, let the l BGC be the greater.
At G, in BG, make the l BGK equal to the l EHF. I. 23.

Proof. In the equal O" ABC, DEF,

because the l BGK = the l EHF, at the centres; Constr,
,', the arc BK = the arc EF. ill. 26.

But the arc BC = the arc EF ; ffyp.

.'. the arc BK = the arc BC,
a part equal to the whole, which is impossible.

.". the L BGC is not unequal to the l EHF;
that is, the l BGC = the l EHF.

And since the z_ BAC at the O"* is half the l BGC at the
centre, ill. 20.

and likewise the l EDF is half the l EHF,

.-. the L. BAC = the i. EDF. Ax. 7.

Q.E.D.
H.S.E. O

210 EUCLID'S ELEMENTS.

Proposition 28. Theorem.

In equal circles the arcs, which are cut off by eqv/il chords,
shall he equal, the major arc equal to the major arc, and the
minor to the minor.

Let ABC, DEF be equal circles ;
and let the chord BC = the chord EF.

Then shall the major arc BAG = the major arc EDF ;
and the minor arc BGC = the minor arc EHF.

Construction.

Find K and L the centres of the 0' ABC, DEF ; ill. 1.
and join BK, KG, EL, LF.

Proof. Because the 0* ABC, DEF are equal,

.-. their radii are equal.

Hence in the A' BKC, ELF,
[ BK = EL,
Because \ KG = LF,

(and BG = EF ; Hyp.

.-. the L BKC = the l ELF. L 8.

.*. the arc BGC = the arc EHF;
for these arcs subtend equal angles at the centre ; ill. 26.

and they are the minor arcs.

But the whole 0** ABGG = the whole 0*'*' DEHF ; Hyp.
.'. the remaining arc BAG = the remaining arc EDF :
and these are the major arcs. q.e.d.

[For Exercises see p. 212.]

BOOK III. PROP. 29.

211

Proposition 29. Theorem.

In equal circles the chordsy which cut off equal arcs, shall
he equal.

Let ABC, DEF be equal circles ;
and let the arc BGC = the arc EHF.

Then shall the chord BC = the chord EF.

Construction. Find K, L the centres of the circles.

Join BK, KC, EL, LF.

Proof. In the equal 0* ABC, DEF,

because the arc BGC = the arc EHF,
.-. the L BKC = the l ELF, at the centres, ill. 27.

Hence in the A' BKC, ^LF,

BK = EL, being radii of equal circles ;
KC =^ LF, for the same reason,
and the l BKC = the l ELF ; Proved.

.'. BC = EF. I. 4.

Q.E.D.

Because^

EXERCISES.

ON PROPOSITIONS 26, 27.

1. 7/ two chords of a circle are parallel^ they intercept equal arcs.

2. The straight lines, which join the extremities of two equal
arcs of a circle towards the same parts, are parallel.

3. In a circle^ or in equal circles, sectors are equal if their angles
at the centres are equal.

212 Euclid's elements.

4. If two chords of a circle intersect at right angles, the opposite
arcs are together equal to a semi-circumference.

5. If two chords intersect toithin a circhy they form an angle
eqiml to that subtended at the circumference by the sum of the arcs
they cut off.

6. If two chords intersect without a circle, they form an angle
equcd to that subtended at the circumference by the difference of the
arcs they cut off.

7. If AB is a fixed chord of a circle, and P any point on one of the
arcs cut off by it, then the bisector of the angle APB ciUs the conjugate
arc in the same point, whatever be the position of P.

8. Two circles intersect at A and B ; and through these points
straight lines are drawn from any point P on the circumference of
one of the circles : shew that when produced they intercept on the
other circumference an arc which is constant for all positions of P.

9. A triangle ABC is inscribed in a circle, and the bisectors of
the angles meet the circumference at X, Y, Z. Find each angle of
the triangle XYZ in terms of those of the original triangle.

ON PROPOSITIONS 28, 29.

10. The straight lines which join the extremities of parallel
chords in a circle (i) towards the same parts, (ii) towards opposite
parts, are equal.

11. Through A, a point of intersection of two equal circles, two
straight lines rAQ, A AY are drawn : shew that the chord PX is
equal to the chord QY.

12. Through the points of intersection of two circles two parallel
straight lines are drawn terminated by the circumferences : shew
that the straight lines which join their extremities towards the same
parts are equal.

13. Two equal circles intersect at A and B ; and through A any
straight line PAQ is drawn terminated by the circumferences : shew
thatBP=BQ.

14. ABC is an isosceles triangle inscribed in a circle, and the
bisectors of the base angles meet the circumference at X and Y.
Shew that the figure BXAVC must have four of its sides equal.

What relation must subsist among the angles of the triangle ABC,
in order that the figure BXAYC may be equilateral ?

NOTE ON BOOK III., PROPS. 26, 27.

213

Note. We have given Euclid's demonstrations of Propositions
26, 27 ; but it should oe noticed that these propositions also admit
of proof by the method of superposition.

To illustrate this method we will apply it to Proposition 26.

Proposition 26. [Alternative Proof.]

In equal circles, the arcs which subtend equal angles, whether at
the centres or circumferences, shall he equal.

Let ABC, DEF be equal circles ;
and let the Z." BGC, EHF at the centres be equal,
and consequently the L^ BAG, EDF at the O**" equal, in. 20.
Then shall the arc BKC= the arc ELF.

Proof. For if the ABG be applied to the DEF, so that the
centre G may fall on the centre H,

then because the circles are equal, ffyp-

.'. their O'*' must coincide ;
hence by revolving the upper circle about its centre, the lower circle
remaining fixed,

B may be made to coincide with E,
and consequently GB with HE.

And because the Z-BGC = the Z. EHF, Jlyp-

.'. GC must coincide with H F :

an d since GC = H F , -ffyP-

:. C must fall on F.

Now B coincides with E, and with F,

and the 0«« of the ABC with the O^ of the DEF ;

.*. the arc BKC must coincide with the arc ELF.

.'. the arc BKC = the arc ELF.

Q.E.D.

214 Euclid's elements.

Proposition 30. Problem.
To bisect a given arc.

Let ADB be the given arc.
It is required to bisect the arc ADB.

Construction. Join AB; and bisect AB at C. I. 10.

At C draw CD at rt. angles to AB, meeting the given
arc at D. I. 11.

Then shall the arc ADB be bisected at D.
Join AD, BD.

Because '

Proof. In the A* ACD, BCD,

AG = BC, Constr.

and CD is common ;
and the l ACD = the l BCD, being rt. angles :

.-. AD = BD. L 4.

And since in the ADB, the chords AD, BD are equal,
.*. the arcs cut off by them are equal, the minor arc equal
to the minor, and the major arc to the major : lii. 28.
and the arcs AD, BD are both minor arcs,
for each is less than a semi-circumference, since DC, bisecting
the chord AB at rt. angles, must pass through the centre
of the circle. iii. 1. Cai\

.-. the arc AD = the arc BD :
that is, the arc ADB is bisected at D. Q.E,F.

EXERCISES.

1. If a tangent to a circle is parallel to a chord, the point of
contact will bisect the arc cut off by the chord.

2. Trisect a quadrant, or the fourth part of the circumference,
of a circle.

BOOK III. PROP. 30. 215

Note. The following alternative proof of Proposition 30 re-
moves the necessity of distinguishing between the major and minor
arcs cut off by the chords AD, BD.

Proposition 30. [Alternative Proof.]

The construction being made as before, we may proceed
thus :

Proof. In the A' ACD, BCD,

{AC = BC, Cmistr.

and CD is common ;
and the l ACD = the l BCD, being rt. angles :

.-. the L. DAC = the l DBC : I. 4.

that is, the l DAB = the z_ DBA.

But these are angles at the O**® subtended by the arcs
DB, DA;

.-. the arc DB = the arc DA: ill. 26.

that is, the arc ADB is bisected at D. Q.E.F.

QUESTIONS for REVISION.

1. When is a straight line said (i) to meet, (ii) to cut, (iii) to
touchy the circumference of a circle ?

2. When are circles said to touch one another? Distinguish
between internal and external contact.

3. What theorems have been so far proved by Euclid regarding
(i) circles which cut one another, (ii) circles which touch one another ?

4. If two unequal circles are concentric, shew that one must lie
wholly within the other.

5. Shew how to divide tKe circumference of a circle into three,
four, or six equal parts.

6. Enunciate the propositions so far proved by Euclid relating
to the properties of a tangent to a circle.

216 euclid's elements.

Proposition 31. Theorem.

The angle in a semicircle is a right angle.

The angle in a segment greater than a semicircle is less than
a right angle.

The angle in a segmsnt less than a semicircle is greater than
a right angle,

Let ABCD be a circle, of which BC is a diameter, and E
the centre ; and let AC be a chord dividing the circle into
the segments ABC, ADC, of which the segment ABC is greater,
and the segment ADC is less than a semicircle.

Then (i) the angle in the semicircle BAC slmll he a right angle ;

(ii) the angle in the segment ABC shall he less than a rt.
angle ;

(iii) the angle in the segment ADC shall he greater than a rt.
angle.

Construction. In the arc ADC take any point D ;
Join BA, AD, DC, AE ; and produce BA to F.

Proot (i) Because EA=EB, L Def. 15.

.-. the A EAB = the l EBA. L 5.

And because EA = EC,

.*. the L EAC = the l ECA.
.'. the whole l BAC = the sum of the /.' EBA, ECA :
but the ext. l FAC = the sum of the two int. /.' CBA, BCA;

.*. the L BAC = the L FAC ;

.'. these angles, being adjacent, are rt. angles.

,*. the L BAC, in the semicircle BAC, is a rt. angle.

BOOK III. PROP. 31. 217

(ii) In the A ABC, because the sum of the /.* ABC, BAC

is less than two rt. angles • I. 17.

and of these, the l BAC is a rt. angle ; Proved.

.". the L ABC, which is the angle in the segment ABC, is

less than a rt. angle.

(iii) Because ABCD is a quadrilateral inscribed in the
O ABC,
/. the opp. Z-'ABC, ADC together = two rt. angles; ill, 22.

and of these, the l ABC is less than a rt. angle : Proved.
.•. the L ADC, which is the angle in the segment ADC, is

greater than a rt. angle. Q.E.D.

EXERCISES.

1. A circle described on the hypotenuse of a right-angled trixangle
as diameter^ passes through the opposite angular point,

2. A system of right-angled triangles is described upon a given
straight line as hypotenuse ; find the locus of the opposite angular
point€i.

3. A straight rod of given length slides between two straight
rulers placed at right angles to one another; find the locus of its
middle point.

4. Two circles intersect at A and B ; and through A two
diameters AP, AQ are drawn, one in each circle : shew that the
points P, B, Q are collinear. [See Def. p. 110.]

5. A circle is described on one of the equal sides of an isosceles
triangle as diameter. Shew that it passes through the middle point
of the base.

6. Of two circles which have internal contact, the diameter of
the inner is equal to the radius of the outer. Shew that any chord
of the outer circle, drawn from the point of contact, is bisected by
the circumference of the inner circle.

7. Circles described on any two sides of a triangle as diameters
intersect on the third side, or the third side produced.

8. Find the locus of the middle points of chords of a circle drawn
through a fixed point. Distinguish between the cases when the given
point is within^ on^ or without the circumference.

9. Describe a square equal to the diflference of two given squares.

10. Through one of the points of intersection of two circles
draw a chord of one circle which shall be bisected by the other.

11. On a given straight line as base a system of equilateral
four-sided figures is described : find the locus of the intersection of
their diagonals.

218

EUCLID'S ELEMENTS.

NOTES ON PROPOSITION 31.

Note L The extension of Proposition 20 to straight and reflex
angles furnishes a simple alternative proof of
the first theorem contained in Proposition 31,
namely,

The angle in a semicircle is a right angle.

For, in the adjoining figure, the angle at
the centre, standing on the arc BHC, is
double the angle BAC at the O^, standing
on the same arc.

Now the angle at the centre is the straight angle BEC ;

/. the L BAC is half of the straight angle BEC *.

and a straight angle = two rt. angles ;

.*. the L BA5 = one half of two rt. angles,

= one rt. angle. Q.B.D.

Note 2. From Proposition 31 we may derive a simple practical
solution of Proposition 17, namely,

To draw a tangent to a circle from a given external point.

Let BCD be the given
circle, and A the given exter-
nal point.

It is required to draw from
A a tangent to the BCD.

Find E, the centre of the
given circle, and join AE.

On AE describe the semi-
circle ABE, to cut the given
circle at B.

Join AB.
Then AB shall he a tangent
to the O BCD.

For the L ABE, being in a semicircle, is a rt. angle. xii. 31.

.'. AB is drawn at rt. angles to the radius EB, from its ex-
tremity B ;

.'. AB is a tangent to the circle. iii. 16.

Q. E. F.

Since the semicircle might be described on either side of AE, it
is clear that there will be a second solution of the problem, as shewn
by the dotted lines of the figure.

BOOK III. QUESTIONS FOR REVISION. 219

QUESTIONS FOR REVISION AND NUMERICAL EXERCISES.

1. Define an arCt a chord, a segment of a circle. When are
segments of circles said to be similar to one another ?

2. Enunciate propositions which give the properties of chords of
a circle in relation to the centre.

3. Prove that in a circle whose diameter is 34 inches, a chord
30 inches in length is at a distance of 8 inches from the centre.

4. In a circle a chord 2 feet in length stands at a distance of
5 inches from the centre : shew that the diameter of the circle is
2 inches longer than the chord.

5. What must be the length of a chord which is 1 foot distant
from the centre of a circle, if the diameter is 2 yards 2 inches ?

6. Two parallel chords of a circle, whose diameter is 13 inches,
are respectively 5 inches and 1 foot in length : shew that the
distance between them is 8^ inches, or 3^ inches.

7. Two circles, whose radii are respectively 26 inches and 25
inches, intersect at two points which are 4 feet apart. Shew that
the distance between their centres is 1 7 inches.

8. The diameters of two concentric circles are respectively
50 inches and 48 inches : shew that any chord of the outer circle
which touches the inner must be 14 inches in length.

9. Of two concentric circles the diameter of the greater is
74 inches, and any chord of it which touches the smaller circle is
70 inches in length : shew that the diameter of the smaller circle
is 2 feet.

10. Two circles of diameters 74 and 40 inches respectively have
a common chord 2 feet in length : shew that the distance between
their centres is 51 inches.

11. The chord of an arc is 24 inches in length, and the height of
the arc is 8 inches ; shew that the diameter of the circle is 26 inches.

12. AB is a line 20 inches in length, and C is its middle point.
On AB, AC, CB semicircles are described. Shew that if a circle is
inscribed in the space enclosed by the three semicircles its radius
must be 3^ inches.

220

EUCLID'S ELEMENTS.

Proposition 32. Theorem.

If a straight line touches a circle, and from the point of
contact a chord is drawn, the angles which this chord makes
with the tangent shall be equal to the angles in the alternate
segments of the circle.

Let EF touch the given 0ABC at B, and let BD be a
chord drawn from B, the point of contact.

Then shall

(i) the L DBF — the angle in the alternate segment BAD :

(ii) the L DB£ = the angle in the alternate segment BCD.

Construction. From B draw BA perp. to EF.
Take any point C in the arc BD ;
and join AD, DC, CB.

L 11.

(i) Proof. Because BA is drawn perp. to the tangent
EF, at its point of contact B,

.-. BA passes through the centre of the circle : ill. 19.

.-. the L ADB, being in a semicircle, is a rt. angle : ill. 31.

.-. in the A ABD, the other z_" ABD, BAD together = a rt.

angle ; L 32.

that is, the ^^ ABD, BAD together = the L ABF.

From these equals take the common l ABD ;
.•. the L DBF = the l. BAD, which is in the alternate seg-
ment.

BOOK III. PROP. 32. 221

(ii) Because ABCD is a quadrilateral inscribed in a
circle,

.\ the opp. z." BCD, BAD together = two rt. angles : ill. 22.

but the iJ DBE, DBF together = two rt. angles ; I. 13.

.-. the z_" DBE, DBF together = the i^ BCD, BAD;

and of these the l DBF = the l BAD ; Proved.
.*. the L DBE = the l BCD, which is in the alternate segment.

Q.E.D.

EXERCISES.

1. State and prove the converse of Proposition 32.

2. Use this proposition to shew that the tangents drawn to a
circle from an external point are equal

3. If two circles touch one another, any straight line drawn
through the point of contact cuts oflF similar segments.

Prove this for (i) internal, (ii) external contact.

4. If two circles touch one another, and from A, the point of
contact, two chords APQ, AXY are drawn : then PX and QY are
parallel.

Prove this for (i) internal, (ii) external contact.

5. Two circles intersect at the points A, B: and one of them
passes through O, the centre of the other : prove that OA bisects
the angle between the common chord and the tangent to the first
circle at A.

6. Two circles intersect at A and B ; and through P, any point
on the circumference of one of them, straight lines PAC, PdD are
drawn to cut the other circle at C and D : shew that CD is parallel
to the tangent at P.

7. If from the point of contact of a tangent to a circle, a chord
is drawn, the perpendiculars dropped on the tangent and chord from
the middle point of either arc cut oflF by the chord are equal.

222

Euclid's elements.

Proposition 33. Problem.

On a given straight line to describe a segment of a circle
which shall contain an angle equal to a given angle,

H H

Let AB be the given st. line, and C the given angle.
It is required to describe on AB a segment of a circle which shall
contain an angle equal to C.
Construction.

At A in BA, make the l BAD equal to the l C. I. 23.
From A draw AE at rt. angles to AD. I. II.

Bisect AB at F. I. 10.

From F draw FG at rt. angles to AB, cutting AE at G.

Join GB.
Then in the A" AFG, BFG,

fAF = BF, Constr.

and FG is common,
and the l AFG = the l BFG, being rt. angles ;
.-. GA = GB: L 4.

.-. the circle described with centre G, and radius GA, will
pass through B.

Describe this circle, and call it ABH.
Then the segment AHB shall contain an angle equal to C.

Proof. Because AD is drawn at rt. angles to the radius
GA from its extremity A,

.-. AD is a tangent to the circle ; III. 16.

and from A, its point of contact, a chord AB is drawn ;
.*. the L BAD = the angle in the alt. segment AHB. in. 32.

But the L BAD == the z_ C : Constr.

,', the angle in the segment AHB = the z. C.

.', AHB is the segment required. Q.E.F.

BOOK III. PROP. 33. 223

Note. In the particular case
when the given angle C is a rt. angle,
the segment required will be the
semicircle described on the given st.
line AB ; for the angle in a semi-
circle is a rt. angle. iii. 31.

EXERCISES.

[The following exercises depend on the corollary to the Converse
of Proposition 21 given on page 201, namely

The locu8 of the vertices of triangles which stand on the same hcLse
and have a given vertical amjle, is the arc of the segment standing on
this base, and containing an angle equal to the given angle.

Exercises 1 and 2 afford good illustrations of the method of
finding required points by the Intersection of Loci. See page 125.]

1. Describe a triangle on a given base, having a given vertical
angle f and having its vertex on a given straight line.

2. Construct a triangle, having given the base, the vertical angle
and (i) one other side.

(ii) the aZtitvde.

(iii) the length of the median which bisects the base.

(iv) the point at which the perpendicular from the vertex meets
the base.

3. Construct a triangle having given the base, the vertical angle,
and the point at which the base is cut by the bisector of the vertical
angle,

[Let AB be the base, X the given point in it, and K the given
angle. On AB describe a segment of a circle containing an ancle
equal to K ; complete the O** by drawing the arc APB. Bisect the
arc APB at P : join PX, and produce it to meet the O** at C. Then
ABC shall be the required triangle.]

4. Constr^uU a triangle hamng given the base, the vertical angle,
and the sum of the remaining sides.

[Let AB be the given base, K the given angle, and H the given
line equal to the sum of the sides. On AB describe a segment
containing an angle etjual to K, also another segment containing
an angle equal to half the L K. From centre A, with radius H,
describe a circle cutting the arc of the last drawn segment at X and
Y. Join AX (or AY) cutting the arc of the first segment at C. Then
ABC shall be the required triangle. ]

5. Construct a triangle having given the base, the vertical angle,
and the difference of the remaining sides.

224 Euclid's elements.

Proposition 34. Problem.

From a given circle to cut off a segment which shall contain
an angle equal to a given angle.

E B F

Let ABC be the given circle, and D the given angle.

It is required to cut off from the O ABC a segment which shall
contain an angle equal to D.

Construction. Take any point B on the O**,

and at B draw the tangent EBF. IIL 17.

At B, in FB, make the l FBC equal to the l D. I. 23.

Then the segment BAC shall contain an angle equal to D.

Proof. Because EF is a tangent to the circle, and from
B, its point of contact, a chord BC is drawn,
.*. the L FBC = the angle in the alternate segment BAC.

in. 32.
But the L FBC = the l D ; Constr,

,', the angle in the segment BAC = the l D.
Hence from the given O ABC a segment BAC has been
cut off, containing an angle equal to D. Q.E.F.

exercises.

1. The chord of a given segment of a circle is produced to a
fixed point : on this straight line so produced draw a segment of a
circle similar to the given segment.

2. Through a given point without a circle draw a straight line
that will cut off a segment capable of containing an angle equal to a
given angle.

QUESTIONS FOR REVISION. 225

QUESTIONS FOR REVISION.

1. Enunciate the propositions from which we infer that a
straight line and a circle must either

(i) intersect in two points ; or
(ii) touch at one point ; or
(iii) have no point in common.

2. Give two independent constructions for drawing a tangent
to a circle from an external point.

Shew that the two tangents so drawn

(i) are equal ;

(ii) subtend equal angles at the centre ;
(iii) make equal angles with the straight line which joins the
given point to the centre.

3. Enunciate propositions relating to

(i) angles in a segment of a circle ;
(ii) similar segments of circles.

4. What are conjugate arcs of a circle ?

The angles in conjugate segments of a circle are supplementary.
How does Euclid enunciate this theorem? State and prove its
converse.

5. Explain what is meant by a reflex angle. What simplifica-
tions may be made in the proofs of Third Book Propositions if reflex
angles are admitted ?

6. K the circumference of a circle is divided into six equal arcs,
shew that the chords joining successive points of division are all
equal to the radius of the circle.

7. Find the locus of the centres of all circles

(i) which pass through two given points ;
(ii) which touch a given circle at a given point ;
(iii) which are of given radius, and touch a given circle ;
(iv) which are of given radius, and pass through a given

point ;
(v) which touch a given straight line at a given point ;
(vi) which touch each of two parallel straight lines ;
(vii) which touch each of two intersecting straight lines of
unlimited length.

8. If a system of triangles stand on the same base and on the
same side of it, and have equal vertical angles, shew that the locus
of their vertices is the arc of a circle. Prove this theorem, having
first enunciated the proposition of which it is the converse.

H.S.B. P

226 euclid's elements.

Proposition 35. Theorem.

If two chords of a circle cut one another, the rectangle
contained by the segments of one shall he equal to the rectangle
contained by the segments of the other.

Let AB, CD, two chords of the ACBD, cut one another
at E.

Then shall the red, AE, EB = the rect, CE, ED.

Construction. Find F, the centre of the ACB ; IIL 1.
From F draw FG, FH perp. respectively to AB, CD. L 12.

Join FA, FE, FD.

Proof. Because FG is drawn from the centre F perp. to AB,

.-. AB is bisected at G. m, 3.

For a similar reason CD is bisected at H.
Again, because AB is divided equally at G, and unequally at E,
.-. the rect. AE, EB with the sq. on EG = the sq. on AG, II. 5.

To each of these equals add the sq. on GF ;
then the rect. AE, EB with the sqq. on EG, GF = the sum of
the sqq. on AG, GF.

But the sqq. on EG, GF = the sq. on FE ; L 47.
and the sqq, on AG, GF = the sq. on AF ;
for the angles, at G are rt. angles.
.-. the rect. AE, EB with the sq. on FE = the sq. on AF.
Similarly it may be shewn that
the rect. CE, ED with the sq. on FE = the sq. on FD.
But the sq. on AF = the sq. on FD; for AF= FD.
.*. the rect. AE, EB with the sq. on FE = the rect. CE, ED
with the sq. on FE.

From these equals take the sq. on FE :
then the rect. AE, EB==the rect. CE, ED. Q.E.D.

BOOK III. PROP. 35. 227

Corollary. If through a fixed point within a circle any
number of cho)'ds are drawn, the rectangles contained by their
segments are all equal.

Note. The following special cases of this proposition deserve
notice :

(i) when the given chords both pass through the centre ;
(ii) when one chord passes through the centre, and cuts the

other at right angles :
(iii) when one chord passes through the centre, and cuts the

other obliquely.

In each of these cases the general proof requires some modifica-
tion, which may be left as an exercise to the student. •

EXERCISES.

1. Two straight lines AB, CD intersect at E, so that the rectangle
AE, EB is equal to the rectangle CE, ED ; shew that the four points
A, B, C, D are concyclic.

2. The rectangle contained by the segments of any chord drawn
through a given point within a circle is equal to the square on half
the shortest chord which may be drawn through that point.

3. ABC is a triangle right-angled at C ; and from C a perpen-
dicular CD is drawn to the hypotenuse : shew that the square on
CD is equal to the rectangle Au, DB.

4. ABC is a triangle ; and AP, BQ, the perpendiculars dropped
from A and B on the opposite sides, intersect at O : shew that the
rectangle AG, OP is equal to the rectangle BO, OQ.

5. Two circles intersect at A and B, and through any point in
AB their common chord two chords are drawn, one in each circle ;
shew that their four extremities are concyclic.

6. A and B are two points within a circle such that the rectangle
contained by the segments of any chord drawn through A is equal to
the rectanele contained by the segments of any chord through B :
shew that A and B are equidistant from the centre.

7. ff through E, a point without a circle^ two secants, EAB, ECD
are drawn; shew thai the rectangle EA, EB is equal to the rectangle
EC, ED.

[Proceed as in iii. 35, using ii. 6.]

8. Through A, a point of intersection of two circles, two straight
lines CAE, DAF are drawn, each passing through a centre and
terminated by the circumferences : shew that the rectangle CA, AE
is equal to the rectangle DA, AF.

228 euclid's elements.

Proposition 36. Theorem.

If from any point without a circle a tangent and a secant
are drawn, then the rectangle contained by the whole secant and
the part of it mthout the circle shall be equal to the square on
the tangent.

Let ABC be a circle ; and from D, a point without it, let
there be drawn the secant DCA, and the tangent DB.

Then the red. DA, DC shall be equal to the sq, on DB.

Construction. Find E, the centre of the ABC : in. 1.
and from E, draw EF perp. to AD. l. 12.

Join EB, EC, ED.

Proof. Because EF, passing through the centre, is
perp. to the chord AC,

.*. AC is bisected at F. ill. 3.

And since AC is bisected at F and produced to D,
.•. the rect. DA, DC with the sq. on FC = the 9q. on FD. IL 6.

To each of these equals add the sq. on EF :
then the rect. DA, DC with the sqq. on EF, FC = the sqq. on

EF, FD.
But the sqq. on EF, FC = the sq. on EC; for EFC is a rt. angle;

= the sq. on EB.
And the sqq. on EF, FD = the sq. on ED; for EFD is a rt. angle;

= the sqq. on EB, BD; for EBD is a
rt. angle. in. 18.

.'. the rect. DA, DC with the sq. on EB = the sqq. on EB, BD.
Fronx these equals take the sq. on EB :
then the rect. DA, DC = the sq. on DB. Q.E.D.

Note. This proof may easily be adapted to the case when the
secant passes through the centre of the circle.

BOOK III. PROP. 36. 229

Corollary. If from a given point without a circle any
rvumber of secants are dravm, the rectangles contained by the
uhcle secants and the parts of them without the circle are all
equal ; for each of these rectangles is equal to the square on the
tangent drawn front the given point to the circle.

For instance, in the adjoining figure,
each of the rectangles PB, PA and PD, PC
and PF, PE is equal to the square on the
tangent PQ :

.*. the rect. PB, PA

= the rect PD, PC

= the rect. PF, PE.

Note. Remembering that the segments into which the chord AB
is divided at P, are the lines PA, PB, (see Def., page 139) we are
enabled to include the corollaries of Propositions 35 and 36 in a
single enunciation.

If any number of chords of a circle are draivn through a given
point vjtthin or without a circle, the rectangles contained by the
segments of the chords are equal.

EXERCISES.

1. Use this proposition to shew that tangents drawn to a circle
from an external point are equal.

2. If two circles intersect, tangents drawn to them from any
point in their common chord produced are equal.

3. If two circles intersect at A and B, and PQ is a tangent to
both circles ; shew that AB produced bisects PQ.

4* If P is any point on the straight line AB produced, shew that
the tangents drawn from P to all circles which pass through A and
B are equal.

5. ABC is a triangle right-angled at C, and from any point P in
AC, a perpendicular PQ is drawn to the hypotenuse : shew that the
rectangle AC, AP is equal to the rectangle AB, AQ.

6. ABC is a triangle right-angled at C, and from C a perpen-
dicular CD is drawn to the hypotenuse : shew that the rect. AB, AD
is equal to the square on AC.

230 euclid's elements.

Proposition 37. Theorem.

If from a point without a circle there are drawn two straight
lines, one of which cuts the circle, and the other meets it, a/nd if
the rectangle cordained by the whole line which auis the circle a/nd
the part of it without the circle is equal to the square on the line
which Tneets the circle, then the line which meets the circle shall
he a tangent to it.

Let ABC be a circle; and from D, a point without it,
let there be drawn two st. lines DCA and DB, of which
DCA cuts the circle at C and A, and DB meets it ; and let
the rect. DA, DC = the sq. on DB.

Then shall DBbe a tangeni to the cirde.

Construction. From D draw DE to touch the ABC : III. 17.

let E be the point of contact.
Find the centre F, and join FB, FD, FE. in. 1.

ProQf. Since DCA is a secant, and DE a tangent to the circle,
.-. the rect. DA, DC = the sq. on DE, in. 36.
But, by hypothesis, the rect. DA, DC = the sq. on DB;
.*. the sq. on DE = the sq. on DB;
.-. DE = DB.

Hence in the A" DBF, DEF,

r DB = DE, Proved.

Because \ and BF = EF ; L Def 15.
[and DF is common ;

.-. the L. DBF = the l DEF. L 8.

But DEF is a rt. angle, for DE is a tangent; IIL 18.

.'. DBF is also a rt. angle;

and since BF is a radius,
.-. DB touches the O ABC at the point B. Q.E.D.

NOTE ON THE METHOD OF LIMITS AS APPLIED TO TANGENCY.

Euclid defines a tangent to a circle as a straight line which meets
the circumference, but being produced^ does not ctit it : and from this
lefinition he deduces the fundamental theorem that a tangent is per-
pendicular to the radius drawn to the point of contact, ill. Prop. 16.

But this result may also be established by the Method of Limits,
which regards the tangent as the ultimoUe position of a secant when
its ttoo points of intersection with the circumference are brought into
coincidence [See Note on page 165] : and it may be shewn that every
theorem relating to the tangent may be derived from some more
general proposition relating to the secant, by considering the ultimate
case when the two points of intersection coincide.

1. To prove by the Method of Limits that a tangent to a circle
is at right aTigles to the radius drawn to the point of cornAact.

Let ABD be a circle, whose centre
is C ; and PABQ a secant cutting the
0«* in A and B ; and let P'AQ' be the
limiting position of PQ when the point
B is brought into coincidence with A.
Then shall CA be perp, to P'Q'.

Bisect AB at E and join CE :

then CE is perp. to PQ. iii. 3.

Now let the secant PABQ change
its position in such a way that while the
point A remains fixed, the point B con-
tinually approaches A, and ultimately
coincides with it ;

then, however near B ojpproax^hes to A, the st. line CE is always
perp. to PQ, since it joins the centre to the middle point of the
chord AB.

But in the limiting position, when B coincides with A, and the
secant PQ becomes the tangent P'Q', it is clear that the point E
will also coincide with A ; and the perpendicular CE becomes the
radius CA. Hence CA is perp. to the tangent P'Q' at its point of
contact A. q.e.d.

Note. It follows from Proposition 2 that a straight line cannot
cut the circumference of a circle at more than ttoo points. Now when
the two points in which a secant cuts a circle move towards coinci-
dence, the secant ultimately becomes a tangent to the circle ; we
infer therefore that a tangent cannot meet a circle otherwise than
at its point of contact. Thus Euclid's definition of a tangent may
be deduced from that given by the Method of Limits.

232 EUCLID'S ELEMENTS.

2. By this method Proposition 32 vnay he derived as a special
case from Proposition 21.

For let A and B be two points on the O**
of the ABC ;

and let BCA, BPA be any two angles in
the segment BCPA :

then the L BPA = the L BCA. iii. 21.

Produce PA to Q.
Now let the point P continually approach
the fixed point A, and ultimately coincide
with it ;
then, however near P may approach to A,
the L BPQ=the L BCA. ill. 21.

But in the limiting position when Q' Of
P coincides with A,

and the secant PAQ becomes the tangent AQ',
it is clear that BP will coincide with BA,
and the L BPQ becomes the L BAQ'.
Hence the L BAQ' = the L BCA, in the alternate segment. Q.E.D.

The contact of circles may be treated in a similar manner by
adopting the following definition.

Definition. K one or other of two intersecting circles alters its
position in such a way that the two points of intersection continually
approach one another, and ultimately coincide ; in the limiting
position they are said to touch one another, and the point in which
the two points of intersection ultimately coincide is called the
point of contact.

EXAMPLES ON LIMITS.

1. Deduce Proposition 19 from the Corollary of Proposition 1
and Proposition 3.

2. Deduce Propositions 11 and 12 from Ex. 1, page 171.

3. Deduce Proposition 6 from Proposition 5.

4. Deduce Proposition 13 from Proposition 10.

5. Shew that a straight line cuts a circle in two diflerent points,
two coincident points, or not at all, according as its distance from
the centre is less than, equal to, or greater than a radius.

6. Deduce Proposition 32 from Ex. 3, page 202.

7. Deduce Proposition 36 from Ex. 7, page 227.

8. The angle in a semi-circle is a right angle.

To what Theorem is this stat-ement reduced, when the vertex of
the right angle is brought into coincidence with an extremity of the
diameter ?

9. From Ex. 1, page 204, deduce the corresponding property of
a triangle inscribed in a circle.

THEOREMS AND EXAMPLES ON BOOK III. 233

THEOREMS AND EXAMPLES ON BOOK III.

I. ON THE CENTRE AND CHORDS OF A CIRCLE.
[See Propositions 1, 3, 14, 15, 25.]

1. All circles which pass throvgh a fixed pointy and have their
centres on a given straight linCf pass also through a second fixed point.

Let AB be the given st. line, and P the given point.

From P draw PR perp. to AB ;
and produce PR to P', making RP' equal to PR.

Then aU circles which pass through P, and have their centres on
AB, shaU pass also through P.

For let C be the centre of any one of these circles.

Join CP, CP.

Then in the A» CRP, CRP',
r CR is common.

Because-! and RP=RP, Constr.

I and the L CRP = the L CRP', being rt. angles ;

.-. CP=CP; 1.4.

.*. the circle whose centre is C, and which passes through P, must

pass also through P.

But C is the centre of any circle of the system ;
.*. all circles, which pass through P, and have their centres in AB,

pass also through P'. Q.B.D.

2. Describe a circle that shall pass through three given points not
in the same straight line.

234 EUCLID-S ELEMENTS.

3. Describe a circle that shall pass through two given points and
have its centre in a given straight line. When is this impossible ?

4. Describe a circle of given radius to pass through two given
points. When is this impossible ?

5. ABC is an isosceles triangle ; and from the vertex A, as
centre, a circle is described cutting the base, or the base produced,
at X and Y. Shew that BX = CY.

6. If two circles which intersect are cut by a straight line

Parallel to the common chord, shew that the parts of it intercepted
etween the circumferences are equal.

7. K two circles cut one another, any two straight lines drawn
through a point of section, making equal angles with the common
chord, and terminated by the circumferences, are equal.

[Ex. 12, p. 171.]

8. K two circles cut one another, of all straight lines drawn
through a point of section and terminated by the circumferences, the
greatest is that which is parallel to the line joining the centres.

9. Two circles, whose centres are C and D, intersect at A, B ;
and through A a straight line PAQ is drawn tei^inated by the
circumferences : if PC, QD intersect at X, shew that the angle PXQ
is equal to the angle CAD.

10. Through a point of section of two circles which cut one
another draw a straight line terminated by the circumferences and
bisected at the point of section.

11. AB is a fixed diameter of a circle, whose centre is C ; and
from P, any point on the circumference, PQ is drawn perpendicular
to AB ; shew that the bisector of the angle CPQ always intersects
the circle in one or other of two fixed points.

12. Circles are described on the sides of a quadrilateral as
diameters : shew that the common chord of any two consecutive
circles is parallel to the common chord of the other two.

[Ex. 9, p. 105.]

13. Two equal circles touch one another externally, and through
the point of contact two chords are drawn, one in each circle, at
right angles to each other : shew that the straight line joining their
other extremities is equal to the diameter of ei trier circle.

14. Straight lines are drawn from a given external point to the
circumference of a circle : find the locus of their middle points.
[Ex. 3, p. 105.]

15. Two equal segments of circles are described on opposite
sides of the same chord AB ; and through O, the middle point of
AB, any straight line POQ is drawn, intersecting the arcs of the
segments at P and Q : shew that OP=OQ.

THEOREMS AND EXAMPLES ON BOOK III. 235

IL ON THE TANGENT AND THE CONTACT OF CIRCLES.
[See Propositions 11, 12, 16, 17, 18, 19.]

1. AU equal chords placed in a given circle touch a fixed con-
centric circle.

2. If from an external point two tangents are drawn to a circle,
the angle contained by them is double the angle contained by the
chord of contact and the diameter drawn through one of the points
of contact.

3. Two circles touch one another externally, and through the
point of contact a straight line is drawn terminated by the circum-
ferences : shew that the tangents at its extremities are parallel

4. Two circles intersect, and through one point of section any
straight line is drawn terminated by the circumferences : shew that
the angle between the tangents at its extremities is equal to the
angle between the tangents at the point of section.

5. Shew that two parallel tangents to a circle intercept on any
third tangent a segment which subtends a right angle at the centre.

6. Two tangents are drawn to a given circle from a fixed ex-
ternal point A, and any third tangent cuts them produced at P and
Q : shew that PQ subtends a constant angle at the centre of the
circle.

7. In any quadrilateral circuniscrihed about a circle^ the sum of
one pair of opposite sides is equal to the sum of the other pair,

8. If the sum of one pair of opposite sides of a quadrilateral is
equal to the sum of the other pair^ shew thai a circle may he inscribed
in the figure,

[Bisect two adjacent angles of the figure, and so describe a circle to
touch three of its sides. Then prove indirectly by means of the
last exercise that this circle must also touch the fourth side.]

9. Two circles touch one another internally, the centre of the
outer being within the inner circle : shew that of all chords of the
outer circle which touch the inner, the greatest is that which is
perpendicular to the straight line joining the centres.

10. In any triangle, if a circle is described from the middle
point of one side as centre and with a radius equal to half the sum
of the other two sides, it will touch the circles described on these
sides as diameters.

11. Through a given point, draw a straight line to cut a circle,
so that the part intercepted by the circumference may be equal to a
given straight Une.

In order that the problem may be possible, between what limits
must the given line lie, when the given point is (i) without the circle,
(ii) within it 7

236 EUCLID'S ELEMENTS.

12. A series of circles touch a given straight line at a given
point : shew that the tangents to them at the points where they cut
a given parallel straight Ime all touch a fixed circle, whose centre is
the given point.

13. If two circles touch one another internally, and any third
circle be described touching one internally and the other externally ;
then the sum of the distances of the centre- of this, third circle from
the centres of the two given circles is constant.

14. Find the locus of points such that the pairs of tangents
drawn from them to a given circle contain a constsuit angle.

15. Find a point such that the tangents drawn from it to two
given circles may be equal to two given straight lines. When is
this impossible ?

16. If three circles touch one another two and two ; prove that
the tangents drawn to them at the three points of contact are con-
current and equaL

The Common Tangents to Two Cikoles.

17. To draw a common tangent to two circles.

First, When the given circles are external to one another, or
when they intersect.

Let A be the centre of the
greater circle, and B the centre
of the less.

From A, with radius equal
to the diflP* of the radii of the
given circles, describe a circle :
and from B draw BC to touch
the last drawn circle. Join AC,
and produce it to meet the
greater of the given circles at D .

Through b draw the radius BE par^ to AD, and in the same
direction.

Join DE.

Then DE shall he a common tangent to the two given circles.

For since AC = the difif** between AD and BE, Constr.

.'. CD = BE:

and CD is par^ to BE ; Constr,

:. DE is equal and par^ to CB. i. 33.

But since BC is a tangent to the circle at C,

.'. the Z. ACB is a rt. angle ; ni. 18.

hence each of the angles at D and E is a rt. angle : i. 29.
.'. DE is a tangent to both circles. Q.E.F.

THEOREMS AND EXAMPLES ON BOOK III. 237

It follows from hypothesis that the point B is outside the circle
•used in the construction :

.*. two tangents such as BC may always be drawn to it from B ;
iftence ttoo common tangents may always be drawn to the given
circles by the above method. These are called the direct common
tangents.

Secondly, When the given circles are external to one another
and do not intersect, two more common tangents may be drawn.

For, from centre A, with a radius equal to the sum of the radii of
the given circles, describe a circle.

From B draw a tangent to this circle ;

and proceed as before, but draw BE in the direction opposite to AD.

It follows from hypothesis that B is external to the circle used
in the construction ;

.*. two tangents may be drawn to it from B.

Hence ttoo more common tangents may be drawn to the given
circles : these will be found to pass between the given circles, and
are called the transverse common tangents.

Thus, in general, four common tangents may be drawn to two
given circles.

The student should investigate for himself the number of common
tangents which may be drawn in the following special cases, noting
in each case where the general construction fails, or is modified : —

(i) When the given circles intersect :

(ii) When the given circles have external contact :

(iii) When the given circles have internal contact :

(iv) When one of the given circles is wholly within the other.

18. Draw the direct common tangents to two equal circles,

19. If the two direct, or the two transverse, common tangents
are drawn to two circles, the parts of the tangents intercepted
between the points of contact are equal.

20. If four common tangents are drawn to two circles external
to one another ; shew that the two direct, and also the two trans-
verse, tangents intersect on the straight line which joins the centres
of the circles.

21. Two given circles have external contact at A, and a direct
common tangent is drawn to touch them at P and Q : shew that PQ
subtends a right angle at the point A.

22. Two circles have external contact at A, and a direct common
tangent is drawn to touch them at P and Q : shew that a circle
described on PQ as diameter is touched at A by the straight line
which joins the centres of the circles.

238 EUCLID'S ELEMENTS.

23. Two circles whose centres are C and C have external contact
at A, and a direct common tangent is drawn to touch them at P
and Q : shew that the bisectors of the angles PCA^ QC'A meet at
right angles in PQ. And if R is the point of intersection of the
bisectors, shew that RA is also a conmion tangent to the circles.

24. Two circles have external contact at A, and a direct common
tangent is drawn to touch them at P and Q : shew that the square
on PQ is equal to the rectangle contained by the diameters of the
circles.

25. Draw a tangent to a given circle, so that the part of it
intercepted by another given circle may be equal to a given straight
line. When is this impossible ?

26. Draw a secant to two given circles, so that the parts of it
intercepted by the circumferences may be equal to two given straight
lines.

Pboblems on Tangkncy.

Ohs, The following exercises are solved by the Method of Inter-
section of Loci, explained on page 125.

The student should begin by making himself familiar with the
following loci.

(i) Tfie locvs of the centres of circles which pass through two given
points,

(ii) The locus of the centres of circles which touch a given straight
line at a given point.

(iii) T/ie locus of the centres of circles which touch a given circle
at a given point.

(iv) The locus of the centres of circles which touch a given straight
lincj and have a given radius.

(v) The loctLs of the centres of circles which touch a given circle^
and have a given raditts.

(vi) The locus of the centres of circles which touch two given
straight lines.

In each exercise the student should investigate the limits and
relations among the data, in order that the problem may be
possible.

27. Describe a circle to touch three given straight lines.

28. Describe a circle to pass through a given point, and touch a
given straight line at a given point.

29. Describe a circle to pass through a given point, and touch a
given circle at a given point.

THEOREMS AND EXAMPLES ON BOOK III.

239

30. Describe a circle of given radius to pass through a given
point, and touch a given straight line.

31. Describe a circle of given radius to touch two given circles.

32. Describe a circle of given radius to touch two given straight
lines.

33. Describe a circle of given radius to touch a given circle and
a given straight line.

34. Describe two circles of given radii to touch one another and
a given straight line, on the same side of it.

35. If a circle touches a given circle and a given straight line,
shew that the points of contact and an extremity of the diameter of
the given circle at right angles to the given line are collinear.

36. To describe a circle to touch a given circle^ and also to touch a
given straight line at a given point.

Let DEB be the given circle,
PQ the given straight line, and A
the given point in it.

R ts required to describe a circle
to touch the DEB, and also to
touch PQ at A.

At A draw AF perp. to PQ : 1. 1 1.
then the centre of the required
circle must lie in AF. iii. 19.

FindC, thecentreof the© DEB,

III. 1.
and draw a diameter BD perp. to

PQ:

join A to one extremity D, cutting

the 0« at E.

Join CE, and produce it to cut AF at F.

Then F shaM be the centre^ and FA the radius of the required circle.

[Supply the proof : and shew that a second solution is obtained
by joining AB, and producing it to meet the O*^.
Also distinguish between the nature of the contact of the circles,
when PQ cuts, touches, or is without the given circle.]

37. Describe a circle to touch a given straight line, and to
touch a given circle at a given point.

38. Describe a circle to touch a given circle, have its centre in a
given straight line, and pass through a given point in that straight
une.

[For other problems of the same class see page 253.]

340 EUCLID'S ELEMENTS.

Orthogonal Cibolbs.

Ukvinition. Circles which intersect at a point, so that the two
taug«»ut8 at that point are at right angles to one another, are said to
bo orlfekocooud, or to cut one another ortliosronaUy.

S9. In two intersecting circles the angle between the tangents
at oixe point of intersection is equal to the angle between the tangents
at the other.

40l If two circles ctU one another orthogonally, the tangent to
eocA circ/e a^ a point of intersection tviU pass through the centre of
the other circle,

41. If two circles cut one another orthogonally y the square on the
distance between their centres is equal to the sum of the squares on
their radii,

42. Find the locus of the centres of all circles which cut a given
circle orthogonally at a given point.

43. Describe a circle to pass through a given point and cut a
given circle orthogonally at a given point.

III. ON ANGLES IN SEGMENTS, AND ANGLES AT THE
CENTRES AND CIRCUMFERENCES OF CIRCLES.

[See Propositions 20, 21, 22 ; 26, 27, 28, 29 ; 31, 32, 33, 34.]

1. If two chords intersect within a circle, they form an angle
equal to that at the centre, subtended by half the sum of the arcs they
cut off.

Let AB and CD be two chords, intersecting
at E within the given ADBC.

Then shall the Z. AEC be equal to the angle at
the centre, subtended by half the sum of the
arcs AC, BD.

Join AD.
Then the ext. L AEC = the sum of the int.
opp. Z.» EDA, EAD ;

that is, the sum of the L" CDA, BAD.

But the Z." CDA, BAD are the angles at the O** subtended by
the arcs AC, BD ;
.*. their sum = half the sum of the angles at the centre subtended by

the same arcs ;
or, the Z. AEC = the angle at the centre subtended by half the sum
of the arcs AC, BD. q.b.d.

THEOREMS AND EXAMPLES ON BOOK III. 241

2. If two chorda when produced intersect outside a circle^ they form
an angle equal to that at the centre subtended by half the difference of
the arcs they cut off,

3. The siun of the arcs cut oflF by two chords of a circle at right
angles to one another is equal to the semi-circumference.

^ 4. AB, AC are any two chords of a circle ; and P, Q are the
middle points of the minor arcs cut off by them : if PQ is joined,
cutting AB and AC at X, Y, shew that AX=AY.

6. If one side of a quadrilateral inscribed in a circle is produced,
the exterior angle is equal to the opposite interior angle,

6. If two circles intersect, and any straight lines are drawn, one
through each point of section, terminated by the circumferences;
shew that the chords which join their extremities towards the same
parts are parallel.

7. ABCD is a quadrilateral inscribed in a circle ; and the opposite
sides AB, DC are produced to meet at P, and CB, DA to meet at Q :
if the circles circumscribed about the triangles PBC, QAB intersect
at R, shew that the points P, R, Q are coUinear.

8. If a circle is described on one of the sides of a right-angled
triangle, then the tangent drawn to it at the point where it cuts the
hypotenuse bisects the other side.

9. Given three points not in the same straight line : shew how
to find any number of points on the circle which passes through
them, without finding the centre.

10. Through any one of three given points not in the same
straight line, draw a tangent to the circle which passes through
them, without finding the centre.

11. Of two circles which intersect at A and B, the circumference
of one passes through the centre of the other : from A any straight
line is drawn to cut the first at C, the second at D ; shew that CB = uD.

12. Two tangents AP, AQ are drawn to a circle, and B is the
middle point of the arc PQ, convex to A. Shew that PB bisects the
angle APQ.

13. Two circles intersect at A and B ; and at A tangents are
drawn, one to each circle, to meet the circumferences at C and D ; if
CB, BD are joined, shew that the triangles ABC, DBA are equiangular
to one another.

14. Two segments of circles are described on the same chord
and on the same side of it ; the extremities of the common chord are
joined to any point on the arc of the exterior segment : shew that
the arc intercepted on the interior segment is constant.

H.aE. Q

242 EUCLID'S ELEMENTS.

15. If a series of triangles are drawn standing on a fixed base,
and having a given verticsil angle, show that the bisectors of the
vertical angles all pass through a fixed point.

16. ABC is a triangle inscribed in a circle, and E the middle
point of the arc subtended by BC on the side remote from A: if
through E a diameter ED is drawn, shew that the angle DEA is half
the dmerence of the angles at B and C. [See Ex. T, p. 109.]

17. If two circles touch each other internally at a point A, any
chord of the exterior circle which touches the interior is divided at
its point of contact into segments which subtend equal angles at A.

18. If two circles touch one another internally, and a straight
line is drawn to cut them, the segments of it intercepted between
the circumferences subtend equal angles at the point of contact.

The Orthockntrb of a Trianolb.

19. The perpendictUars dravm from the vertices of a triangle to
the opposite sides are concurrent.

In the A ABC, let AD, BE be the
P|erp" drawn from A and B to the oppo-
site sides ; and let them intersect at O.
Join CO ; and produce it to meet AB
at F.

It is required to shew that CF is perp.
to AB.

Join DE.

Then, because the Z.» OEC, ODC are
rt. angles, ^VP-

.'. the points O, E, C, D are concyclic :
.'. the L DEC = the L DOC, in the same segment ;
= the vert. opp. L FOA.

Again, because the Z." AEB, ADB are rt. angles, ^UP'

.'. the points A, E, D, B are concyclic :
/. the L DtB = the L DAB, in the same segment.

/. the sum of the L' FOA, FAO = the sum of the L' DEC, DEB

= a rt. angle : ffyP'

.'. the remaining Z. AFO = a rt. angle : i. 32.

that is, CF is perp. to AB.
Hence the three perp" AD, BE, CF meet at the point O. Q.B.D.

[For an Alternative Proof see p. 114.]

THEOREMS AND EXAMPLES ON BOOK III. 243

Definitions.

(i) The intersection of the perpendiculars drawn from the
vertices of a triangle to the opposite sides is called its ortbocentre.

(ii) The triangle formed by joining the feet of the perpendi-
culars is called the pedal or ortbocentrlc triangle.

20. In an octUe-aTigled triangle the perpendiculars draion from
the vertices to the opposite sides bisect the angles of the pedal triangle
through which they pass.

In the acute-angled A ABC, let AD,
BE, CF be the perp" drawn from the
vertices to the opposite sides, meeting
at the orthocentre O ; and let DEF be
the pedal triangle.

Then shaU AD, BE, CF bisect respec-
tively tlie L* FDE, DEF, EFD.

For, as in the last theorem, it may
be shewn that the points O, D, C, E are
concyclic ;

.'. the L ODE = the L OCE, in the same segment.

Similarly the points O, D, B, F are concyclic ;

.'. the L ODF = the L OBF, in the same segment.

But the L OCE = the L OBF, each being the comp* of the L BAC.

.-. the L ODE = the L ODF.

Similarly it may be shewn that the Z.» DEF, EFD are bisected
by BE and CF. q.e.d.

Corollary, (i) Every two sides of the pedal triangle are equally
inclined to that aide of the original triangle in which they meet.

For the L EDC=the comp* of the L ODE

= the comp* of the L OCE
= the L BAC.

Similarly it may be shewn that the L FDB = the L BAC,
.*. the L EDC = the L FDB = the L A.

In like manner it may be proved that

the L DEC=the L FEA = the L B,
and the L DFB = the L EFA=the L 0.

Corollary, (ii) The triangles DEC^ AEF, DSF are equiangular
to one another and to the triangle ABC.

Note. If the angle BAC is obtuse, then the perpendiculars BE, CF
bisect externally the corresponding angles of the pedal triangle.

244 Euclid's elements.

2L In any triangle, if the perpendiculara drawn from the vertices
on the opposite sides are produced to meet the circumscribed circle,
then each side bisects that portion of the line, perpendicular to it which
lies between the orthocentre and the circumference.

Let ABC be a triangle in which the perpen-
diculars AD, BE are drawn, intersecting at O
the orthocentre, and let AD be produced to meet
the O** of the circumscribing circle at G.

Then shall DO = DG.

Join BG.

Then in the two A- OEA, ODB,
the L OEA = the Z. ODB, being rt. ancles ;
and the L EOA=the vert. opp. L DOB ;

.'. the remaining L EAO=the remaining L DBO. I. 32.

But the L CAG=the L CBQ, in the same segment;
.-. the L DBO = the L DBG.

Then in the A» DBO, DBG,

the L DBO = the L DBG, Proved.

Because-^ the L BDO=the L BDG,
and BD is common ;

.-. DO = DG. I. 26.

Q.E.D.

22. In an acute-angled triangle the three sides a/re the external
bisectors of the angles of the pedal triangle : and in an obtuse-angled
triangle the sides covUaining the obtiise angle are the internal bisectors
of the corresponding angles of the pedal triangle,

23. IfO is the orthocentre of the triangle ABC, shew that the
angles BOC, BAG are supplementary.

24. IfOis the orthocentre of the triangle ABC, then any one oj
the four points O, A, B, C is the orthocentre of the triangle whose
vertices are the other three,

25. The three circles which pa^ss through two vertices of a triangle
and its orthocentre are each equal to the circle circumscribed abotU the
triangle.

26. D, E are taken on the circumference of a semicircle described
on a given straight line AB : the chords AD, BE and AE, BD
intersect (produced if necessary) at F and G : shew that FG is
perpendicular to AB.

27. ABCD is a parallelogram ; AE and CE are drawn at right
angles to AB, and CB respectively : shew that ED, if produced, will
be perpendicular to AC.

THEOREMS AND EXAMPLES ON BOOK III. 245

28. ABC is a triangle, O is its orthocentre, and AK a diameter
of the circumscribed circle : shew that BOCK is a parallelogram.

29. The orthocentre of a triangle is joined to the middle point
of the base, and the joining line is produced to meet the circum-
scribed circle : prove that it will meet it at the same point as the
diameter which passes through the vertex.

30. The perpendicular from the vertex of a triangle on the base,
and the straight line joining the orthocentre to the middle point of
the base, are produced to meet the circumscribed circle at P and Q :
shew that PQ is parallel to the base.

31. The distance of each vertex of a triangle from the orthocentre
is double of the perpendicvlar drawn from the centre of the circum-
scribed circle on the opposite side,

32. Three circles are described each passing through the ortho-
centre of a triangle and two of its vertices : shew that the triangle
formed by joining their centres is equal in all respects to the original
triangle.

33. ABC is a triangle inscribed in a circle, and the bisectors of
its angles which intersect at O are produced to meet the circum-
ference in PQR : shew that O is the orthocentre of the triangle PQR.

34. Construct a triangle, having given a vertex, the orthocentre,
and the centre of the circumscribed circle.

Loci.

35. Given the base and vertical angle of a triangle, find the locus
of its orthocentre.

Let BC be the given base, and X the
given angle ; and let BAC be any triangle ^
on the base BC, having its vertical L A XK
equal to the Z. X. / \

Draw the perp" BE, CF, intersecting
at the orthocentre O.

It is required to find the locus of O.

Since the Z-» OF A, OEA are rt. angles,
.'. the points O, F, A, E are concyclic ;
.'. the L FOE is the supplement of the L A : iii. 22,

.*. the vert. opp. L BOC is the supplement of the L A.

But the Z. A is constant, being always equal to the L X ;

.'. its supplement is constant ;

that is, the A BOC has a fixed base, and constant vertical angle ;

hence the locus of its vertex O is the arc of a segment of whicn BC

is the chord. [See Corollary p. 201. 1

246

Euclid's elements.

36. Given the base and vertical angle of a triangle^ find the locus
of the intersection of the bisectors of its angles.

Let BAG be any triangle on the given
base BC, having its vertical angle equal to
the given L X ; and let Al, Bl, CI be the
bisectors of its angles. [See Ex. 2, p. 111.]

It is required to find the locus of the point I.

Denote the angles of the A ABC by
A, B, C ; and let the Z. BIG be denoted
by I.

Then from the ABIC,

(i) I + JB + JC = two rt. angles,

and from the A ABC,

A + B + C = two rt. angles ;

(ii) so that JA + JB + JC = one rt. angle,

.'. , taking the differences of the equals in (i) and (ii),

I - iA=one rt. angle :
or, l=one rt. angle + iA.

But A is constant, being always equal to the L X ;

.'. I is constant :
/. the locus of I is the arc of a segment on the fixed chord BC.

37. Given the base and vertical angle of a triangle, find the locus
of the centroid, thai is, the intersection of the medians.

Let BAC be any triangle on the given
base BC, having its vertical angle equal
to the given angle S ; let the medians
AX, BY, CZ intersect at the centroid G.
[See Ex. 4, p. 113.]

It is required to find the locus of the point G.

Through G draw GP, GQ par^ to AB
and AC respectively.

Then ZG is a third part of ZC ;

Ex. 4, p. 113,
and since GP is par^ to ZB,
.-. BP is a third part of BC.
Similarly QC is a third part of BC ;

.-. P and Q are fixed points.

Now since PG, GQ are par^ respectively to BA, AC,

.-. the L PGQ=the L BAC,
= the L S,
that is, the L PGQ is constant ;
.*. the locus of G is the arc of a segment on the fixed chord PQ.

Note. In this problem the points A and G move on the arcs of
similar segments.

Ex. 19, p. 107.

Constr.
I. 29.

THEOREMS AND EXAMPLES ON BOOK HI. 247

38. Given the base and the vertical angle of a triangle ; find the
locus of the intersection of the bisectors of the exterior base angles.

39. Through the extremities of a given straight line AB any two
parallel straight lines AP, BQ are drawn ; find the locus of the
intersection of the bisectors of the angles PAB, QBA.

40. Find the locus of the middle points of chords of a circle
drawn through a fixed point.

Distinguish between the cases when the given point is within,
on, or without the circumference.

41. Find the locus of the points of contact of tangents drawn
from a fixed point to a system of concentric circles.

42. Find the locus of. the intersection of straight lines which
pass through two fixed points on a circle and intercept on its cir>
cumference an arc of constant length.

43. A and B are two fixed points on the circumference of a
circle, and PQ is any diameter : find the locus of the intersection of
PA and QB.

44. BAC is any triangle described on the fixed base BC and
having a constant vertical angle ; and BA is produced to P, so that
BP is equal to the sum of the sides containing the vertical angle :
find the locus of P.

45. AB is a fixed chord of a circle, and AC is a moveable chord
passing through A : if the parallelogram CB is completed, find the
locus of the intersection of its diagonals.

46. A straight rod PQ slides between two rulers placed at right
angles to one another, and from its extremities PX, QX are drawn
perpendicular to the rulers : find the locus of X.

47. Two circles whose centres are C and D, intersect at A and
B : through A, any straight line PAQ is drawn terminated by the
circumferences ; and PC, QD intersect at X : find the locus of X,
and shew that it passes through B. [Ex. 9, p. 234.]

48. Two circles intersect at A and B, and through P, any point
on the circumference of one of them, two straight lines PA, Pfe are
drawn, and produced if necessary, to cut the other circle at X and
Y : find the locus of the intersection of AY and BX.

49. Two circles intersect at A and B ; HAK is a fixed straight
line drawn through A and terminated by the circumferences, and
PAQ is any other straight line similarly drawn : find the locus of
the intersection of H P and QK.

248 EUCLID'S ELEMENTS.

60. Two segnients of circles are on the same chord AB and on
the same side of it ; and P and Q are any points one on each arc :
find the locus of the intersection of the bisectors of the angles PAQ,
PBQ.

51. Two circles intersect at A and B ; and through A any straight
line PAQ is drawn terminated by the circumferences : find the locus
of the middle point of PQ.

Miscellaneous Examples on Angles in a Cibcle.

52. ABC is a triangle, and circles are drawn through B, C,
cutting the sides in P, Q, P', Q', ... : shew that PQ, PQ' ... are
parallel to one another and to the tangent drawn at A to the circle
circumscribed about the triangle.

53. Two circles intersect at B and C, and from any point A, on
the circumference of one of them, AB, AC are drawn, and produced
if necessary, to meet the other at D and E : shew that DE is parallel
to the tangent at A.

54. A secant PAB and a tangent PT are drawn to a circle from
an external point P ; and the bisector of the angle ATB meets AB at
C : shew that PC is equal to PT.

55. From a point A on the circumference of a circle two chords
AB, AC are drawn, and also the diameter AF : if AB, AC are pro-
duced to meet the tangent at F in D and E, shew that the triangles
ABC, AED are equiangular to one another.

56. O is any point within a triangle ABC, and OD, OE, OF are
drawn perpendicular to BC, CA, AB respectively : shew that the
angle BOC is equal to the sum of the angles BAC, EDF.

57. If two tangents are drawn to a circle from an external
point, shew that they contain an angle equal to the difference of the
angles in the segments cut off by the chord of contact.

58. Two circles intersect, and through a point of section a
straight line is drawn bisecting the angle between the diameters
through that point : shew that this straight line cuts off similar
segments from the two circles.

59. Two equal circles intersect at A and B ; and from centre A,
with any radius less than AB a third circle is described cutting the
given circles on the same side of AB at C and D : shew that the
points B, C, D are coUinear.

60. ABC and A'B'C are two triangles inscribed in a circle, so
that AB, AC are respectively parallel to A'B', A'C : shew that BC
is parallel to B'C.

THEOREMS AND EXAMPLES ON BOOK Itl. 249

61. Two circles intersect at A and B, and through A two straight
lines HAK, PAQ are drawn terminated by the circumferences : if
HP and KQ intersect at X, shew that the points H, B, K, X are
concyclic.

62. Describe a circle touching a given straight line at a given
point, so that tangents drawn to it from two fixed points in the
given line may be parallel. [See Ex. 10, p. 197.]

63. C is the centre of a circle, and CA, CB two fixed radii : if
from any point P on the arc AB perpendiculars PX, PY are drawn
to CA and CB, shew that the distance XY is constant.

64. AB is a chord of a circle, and P any point in its circum-
ference ; PM is drawn perpendicular to AB, and AN is drawn per-
pendicular to the tangent at P : shew that M N is parallel to PB.

65. P is any point on the circumference of a circle of which AB
is a fixed diameter, and PN is drawn perpendicular to AB ; on AN
and BN as diameters circles are described, which are cut by AP, BP
at X and Y : shew that XY is a common tangent to these circles.

66. Upon the same chord and on the same side of it three seg-
ments of circles are described containing respectively a given angle,
its supplement and a right angle : shew that the intercept made by
the two former segments upon any straight line drawn through an
extremity of the given chord is bisected by the latter segment.

67. Two straight lines of indefinite length touch a given circle,
and any chord is drawn so as to be bisected by the chord of contact :
if the former chord is produced, shew that the intercepts between
the circumference and the tangents are equal.

68. Two circles intersect one another : through one of the points
of section draw a straight line of given length terminated by the
circumferences.

69. On the three sides of any triangle equilateral triangles are
described remote from the given triangle : shew that the circles
described about them intersect at a point.

70. On BC, CA, AB the sides of a triangle ABC, any points
P, Q, R are taken ; shew that the circles described about the triangles
AQR, BRP, CPQ meet in a point.

71. Find a point within a triangle at which the sides subtend
equal angles.

72. Describe an equilateral triangle so that its sides may pass
through three given points.

73. Describe a triangle equal in all respects to a given triangle,
and having its sides passing through three given points.

250 EUCLID'S ELEMENTS.

Simson's Line.

74. If from any point on the circumference of the circle drctim-
sci'ibed aboiU a triangle, perpendicidars are drawn to the three sidesy
the feet of these perpendiculars are collinear.

Let P be any point on the O** of the
circle circumscribed about the A ABC ; and
let PD, PE, PF be the perp" drawn from
P to the three sides.

It is required to prove thai the points
D, E, F are collinear.

Join FD and DE :
then FD and DE shall be in the same st.
line.

Join PB, PC.

Because the L* PDB, PFB are rt. angles, ByP'

:. the points P, D, B, F are concyclic :
.*. the L PDF = the L PBF, in the same segment, in. 2L

But since BACP is a quad^ inscribed in a circle, having one of its
sides AB produced to F,

.-. the ext. L PBF = the opp. int. L ACP. Ex. S,p. 202.
/. the L PDF =the L ACP.

To each add the L PDE :
then the ^» PDF, PDE = the L' ECP, PDE.

But since the ^» PDC, PEC are rt. angles,

.*. the points P, D, E, C are concyclic ;

.*. the L* ECP, PDE together = two rt. angles :

.'. the Z." PDF, PDE together = two rt. angles ;

.*. FD and DE are in the same st. line ; I. 14.

that is, the points D, E, F are collinear. q.£.d.

[The line FDE is called the Pedal or Slxnson's Line of the triangle
ABC for the point P ; though the tradition attributing the theorem
to Robert Simson has been recently shaken by the researches of
Dr. J. S. Mackay.]

75. ABC is a triangle inscribed in a circle ; and from any point
P on the circumference PD, PF are drawn perpendicular to BU and
AB: if FD, or FD produced, cuts AC at E, shew that PE is per-
pendicular to AC.

76. Find the locus of a point which moves so that if perpen-
diculars are drawn from it to the sides of a given triangle, their feet
are collinear.

77. ABC and AB'C are two triangles having a common vertical
angle, and the circles circumscribed about them meet again at P ;
shew that the feet of perpendiculars drawn from P to the four lines
AB, AC, BC, B'C are collinear.

THEOREMS AND EXAMPLES ON BOOK III. 261

78. A triangle is inscribed in a circle, and any point P on the
circumference is joined to the orthocentre of the triangle : shew tha£
this joining line is bisected by the pedal of the point P.

IV. ON THE CIRCLE IN CONNECTION WITH RECTANGLES.

[See Propositions 35, 36, 37.]

1. If from any external point P two tangents are drawn to a given
circle whose centre i» 0, and if OP meets the chord of contact at Q ;
then the rectangle OP, 0^t8 equal to the square an the radius.

Let PH, PK be tangents, drawn from
the external point P to the O H AK, whose
centre is 0; and let OP meet HK the
chord of contact at Q, and the O*^ at A.

Then shaU the rect, OP, OCl=the sq. on
OA.

On HP as diameter describe a circle :
this circle must pass through Q, since the
L HQP is a rt. angle. in. 31.

Join OH.

Then since PH is a tangent to the H AK,

.'. the Z. OHP is a rt. an^le.

And since HP is a diameter of the HQP,

.-. OH touches the HQP at H. in. 16.

.*. the rect. OP, OQ=the sq. on OH, in. 36.

= the sq. on OA. Q.E.D.

2. ABC is a triangle, and AD, BE, OF the perpendiculars drawn
from the vertices to the opposite sides, meeting in the orthocentre O :
shew that the rect. AO, OD = the rect. BO, OE = the rect. CO, OF.

3. ABC is a triangle, and AD, BE the perpendiculars drawn
from A and B on the opposite sides : shew that the rectangle CA, CE
is equal to the rectangle CB, CD.

4. ABC is a triangle right-angled at C, and from D, any point
in the hypotenuse AB, a straight line DE is drawn perpendicular to
AB and meeting BC at E : shew that the square on DE is equal to
the difiPerence of the rectangles AP, DB and CE, EB.

5. From an external point P two tangents are drawn to a given
circle whose centre is O, and OP meets the chord of contact at Q :
shew that any circle which passes through the points P, Q will cut
the given circle orthogonally. [See Def. p. 240.]

252 EUCLID'S ELEMENTS.

6. A series of circles pass through two givefn points^ and from a
fixed point in the common chord produced tangents are dravm to aU
the circles: shew that the points of contact lie on a circle which cuts aU
the given circles orthogonally.

7. All circles which pass through a fixed point, and cut a given
circle orthogonally, pa>ss also throvjgh a second fixed point.

8. Find the locus of the centres of all circles which pass through
a given point and cut a given circle orthogonally.

9. Describe a circle to pass through two given points and cut a
given circle orthogonally.

10. A, B, C, D are four points taken in order on a given straight
line : find a point O between B and C such that the rectangle
OA, OB may be equal to the rectangle OC, OD.

11. AB is a fixed diameter of a circle, and CD a fixed straight
line of indefinite length cviting AB or AB produced at right angles ;
any straight line is drawn through A to cut OD atP and the cirde at
Q : shew that the rectangle AP, AQ is constant.

12. AB is a fixed diameter of a circle, and CD a fixed chord at
right angles to AB ; any straight line is drawn through A to cut CD
at P and the circle at Q : shew that the rectangle AP, AQ is equal
to the square on AC.

13. Aw a fixed point, and CD a fixed straight line of indefinite
length ; AP is any straight line dravm through A to meet CD at P ;
and in AP a point Q is taken such that the rectangle AP, AQ ia
constant : find the locus of Q.

14. Two circles intersect orthogonally, and tangents are drawn
from any point on the circumference of one to touch the other :
prove that the first circle passes through the middle point of the
chord of contact of the tangents. [Ex. 1, p. 251.]

15. A semicircle is described on AB as diameter, and any two
chords AC, BD are drawn intersecting at P : shew that

AB2=AC.AP + BD.BP.

16. Two circles intersect at B and C, and the two direct common
tangents AE and DF are drawn : if the common chord is produced
to meet the tangents at G and H, shew that GH2= AE^H- BC*.

17. If from a point P, without a circle, PM is drawn perpen-
dicular to a diameter AB, and also a secant PCD, shew that

PM2=PC.PD + AM.MB.

THEOREMS AND EXAMPLES ON BOOK III. 253

18. Three circles intersect at D, and their other points of inter-
section are A, B, C ; AD cuts the circle BDC at E, and EB, EC cut
the circles ADB, ADC respectively at F and G : show that the
points F, A, G are coUinear, and F, B, C, G concyclic.

19. A semicircle is described on a given diameter BC, and from
B and C any two chords BE, CF are drawn intersecting within
the semicircle at O ; BF and CE are produced to meet at A : shew
that the sum of the squares on AB, AC is equal to twice the square
on the tangent from A together with the square on BC.

20. X and Y are two fixed points in the diameter of a circle
equidistant from the centre C : through X any chord PXQ is drawn,
and its extremities are joined to Y ; shew that the sum of the
squares on the sides of the triangle PYQ is constant. [See p. 161,
£x. 24.]

Problems on Tangbncy.

21. To deacribe a circle to pass through two given points and to
touch a given straight line.

Let A and B be the given points,
and CD the given st. line.

It is required to describe a circle to
pass through A and B and to touch
CD.

Join BA, and produce it to meet
CD at P.

Describe a square equal to the C
rect. PA, PB ; ii. 14.

and from PD (or PC) cut oflP PQ equal to a side of this square.

Through A, B, and Q describe a circle. Ex. 4, p. 171.

Then since the rect. PA, PB=the sq. on PQ,

.-. the ABQ touches CD at Q. iii. 37.

Q.E.F.

Notes, (i) Since PQ may be taken on either side of P, it is
clear that there are in general two solutions of the problem.

(ii) When AB is parallel to the given line CD, the above method
is not applicable. In this case a simple construction follows from
in. 1, (>)r. and in. 16, and it will be found that only one solution
exists.

254

EUCLID'S ELEMENTS.

22. To describe a circle to pass through two given points and to
touch a given circle.

Let A and B be the given
points, and CRP the given
circle.

It is required td describe a
circle to pass throrigh A and
B, and to touch the Q) CRP.

Through A and B de-
scribe any circle to cut the
given circle at P and Q. '***.».''

Join AB, PQ, and pro- D

duce them to meet at D.

From D draw DC to touch the given circle, and let C be the
point of contact.

Then the circle described through A, B, C wUl touch the given circle.

For, from the ABQP, the rect. DA, DB=the rect. DP, DQ :
and from the OPQC, the rect. DP, DQ=the sq. on DC ; ill. 36.

.*. the rect. DA, DB=the sq. on DC :

.-. DC touches the O ABC at C. ra. 37.

But DC touches the PQC at C ; C(yastr.

.*. the ABC touches the given circle, and it passes through the
given points A and B. Q.E.F.

Note, (i) Since two tangents may be drawn from D to the
given circle, it follows that there will be two solutions of the problem.

(ii) The general construction fails when the straight line bisect-
ing AB at right angles passes through the centre of the given circle :
the problem then becomes symmetrical, and the solution is obvious.

23. To describe a circle to pass through a given point and to
touch two given straight lines.

Let P be the given point, and
AB, AC the given straight lines.

It is required to describe a circle
to pass through P and to touch
AB, AC.

Now the centre of every circle
which touches AB and Au must
lie on the bisector of the L BAC.

Ex. 7, p. 197.

Hence draw AE bisecting the
L BAC.

From P draw PK perp. to AE, and produce it to P,
making KP' equal to PK.

THEOREMS AND EXAMPLES ON BOOK III. 255

Then every circle which has its centre in AE, and passes through

P, must also pass through P'. Ex. 1, p. 233.

Hence the problem is now reduced to drawing a circle through

P and P to touch either AC or AB. Ex. 21, p. 253.

Produce P'P to meet AC at S.
Describe a square equal to the rect. SP, SP'; ii. 14.

and cut off oR equal to a side of the square.
Describe a circle throujgh the points P', P, R.
Then since the rect. SP, SP' = the sq. on SR, Constr.
.'. this circle touches AC at K ; iii. 37,

and since its centre is in AE, the bisector of the L BAC,

it may be shewn also to touch AB. q.e.f.

Notes, (i) Since SR may be taken on either side of S, it follows
that there will be two solutions of the problem.

(ii) If the given straight lines are parallel, the centre lies on the
parallel straight line mid- way between them, and the construction
proceeds as before.

24. To describe a circle to touch two given straight lines and a
given circle.

Let AB, AC be the two given Jj'

Bt. lines, and D the centre of the ^''' ^-p

given circle. ^'''' "yr

It is required to describe a circle y'/ ^\^ ^^^ \

to touch AB, AC and the circle Q y' • J^\\/^%^'*\

whose centre is D. ' ; v^oV-4^ ! \

Draw EF, GH pari to AB J^ SkT V^ )
and AC respectively, on the sides A \^ Pj V y / B
remote from D, and at distances ****»«Lr-'^^'*'^
from them equal to the radius £ M F

of the given circle.

Describe the MND to touch EF and QH at M and N, and to
pass through D. Ex. 23, p. 254.

Let O be the centre of this circle.

Join OM, ON, OD meeting AB, AC, and the given circle at P, Q,
and R.

Then a circle described with centre O and radius OP wiU touch
AB, AC and the given circle.

For since O is the centre of the MND,

.-. OM = ON=OD.
But PM = QN = RD; Constr.

.: OP=OQ = OR.
.*. a circle described with centre O, and radius OP, will pass
through Q and R.

And since the L!^ at M and N are rt. angles, iii. 18.

.'. the Z." at P and Q are rt. angles ; i. 29.

.*. the PQR touches AB and AC.

256

EUCLID'S ELEMENTS.

And since R, the point in which the circles meet, is on the line of
centres OD,

/. the PQR touches the given circle. Q.E.F.

KoTE. There will be two solutions of this problem, since two
circles may be drawn to tcuch EF, QH and to pass through D,

25. To describe a circle to pass through a given point and Uyach a
given straight line and a given circle.

Let P be the given point, AB the
given St. line, and DHE the given
circle, of which C is the centre.

It is required to describe a circle to
pass through P, and to touch AB
and the DHE.

Through C draw DCEF perp. to
AB, cutting the circle at the points
D and E, of which E is between C
and AB.

Join DP;
and by describing a circle through

F, E, and P, find a point K in DP (or DP produced) such that tbo
rect. DE, DF=therect. DK, DP.

Describe a circle to pass through P, K, and touch AB: Ex. 21, p. 253«
This circle shall also touch the given DHE.

For let G be the point at which this circle touches AB.
Join DG, cutting the given circle DHE at H.

Join HE.

Then the Z. DHE is a rt. angle, being in a semicircle,
also the angle at F is a rt. angle ;
.'. the points E, F, G, H are concyclic ;
.-. the rect. DE, DF= the rect. DH, DG :
but the rect. DE, DF = the rect. DK, DP:
/. the rect. DH, DG= the rect. DK, DP:
.*. the point H is on the PKG.

Let O be the centre of the PHG.

Join OG, OH, CH.
Then OG and DF are par*, since they are both perp. to AB ;

and DG meets them.
.-. theZ. OGD = the^ GDC. 1.29.

But since OG = OH, and CD = CH,
.-. the L OGH = the L OHG ; and the L CDH =the L CHD :

.-. theZ.OHG=the^CHD;
.*. OH and CH are in one st. line.
.*. the PHG touches the given DHE. q.e.f.>

m. 3L

Constr,

III. 36.
Constr»

THEOREMS AND EXAMPLES ON BOOK III. 257

Notes, (i) Since two circles may be drawn to pass through P, K
and to touch AB, it follows that there will be two solutions of the
present problem.

(ii) Two more solutions may be obtained by joining PE, and
proceeding as before.

The student should examine the nature of the contact between
the circles in each case.

26. Describe a circle to pass through a given point, to touch
a given straight line, and to have its centre on another given
straight line.

27. Describe a circle to pass through a given point, to touch a
given circle, and to have its centre on a given straight line.

28. Describe a circle to pass through two given points, and to
intercept an arc of given length on a given circle.

29. Describe a circle to touch a given circle and a given straight
line at a given point.

30. Describe a circle to touch two given circles and a given
straight line.

V. ON MAXIMA AND MINIMA.

We gather from the Theory of Loci that the position of an angle,
line or figure is capable under suitable conditions of gradual change ;
and it is usually found that change of position involves a correspond-
ing and gradual change of magnitv/ie.

Under these circumstances we may be required to note if any
situations exist at which the magnitude in question, after increas-
ing, begins to decrease ; or after decreasing, to increase : in such
situations the magnitude is said to have reached a MftTrimiim or
a Minimum value ; for in the former case it is greater, and in the
latter case less than in adjacent situations on either side. In the
geometry of the circle and straight line we only meet with such
cases of continuous change as admit of one transition from an in-
creasing to a decreasing state — or vice vers^ — so that in all the
problems with which we have to deal (where a single circle is
mvolved) there can be only one Maximum and one Minimum — the
Maximum being the greatest, and the Minimum being the least
value that the variable magnitude is capable of taking.

258 EUCLID'S ELEMENTS.

Thus a variable geometrical magnitude reaches its maximum or
minimum value at a turning pointy towards which the magnitude
may mount or descend from either side : it is natural therefore to
expect a maximum or minimum value to occur when, in the course of
its change, the magnitude assumes a symmetrical form or position ;
and this is usually found to be the case.

This general connection between a symmetrical form or position
and a maximum or minimum value is not exact enough to constitute
B, proof in any particular problem ; but by means of it a situation is
suggested, which on further examination may be shewn to give the
maximum or minimum value sought for.

For example, suppose it is required

to determine the greatest straight line that may he drawn perpendicular
to the chord of a segment of a circle and intercepted between the chord
and the arc :

we immediately anticipate that the greatest perpendicular is that
which occupies a symmetrical position in the figure, namely the
perpendicular which passes through the middle point of the chord ;
and on further examination this may be proved to be the case by
means of i. 19, and i. 34.

Again we are able to find at what point a geometrical magnitude,
varying under certain conditions, assumes its Maximum or Minimum
value, if we can discover a construction for drawing the magnitude
so that it may have an assigned value : for we may then examine
between what limits the assigned value must lie in order that the
construction may be possible ; and the higher or lower limit will
give the Maximum or Minimum sought for.

It was pointed out in the chapter on the Intersection of Loci,
[see page 125] that if under certain conditions existing among the
data, tvx) solutions of a problem are possible, and under other con-
ditions, no solution exists, there will always be some intermediate
condition under which one and only one distinct solution is possible.

Under these circumstances this single or limiting solution will
always be found to correspond to the maximum or minimum value
of the magnitude to be constructed.

1. For example, suppose it is required

to divide a given straight line so that the rectangle contained by the
two segments may be a maocimum.

We may first attempt to divide the given straight line so that
the rectangle contained by its segments may have a given area — that
is, be equsil to the square on a given straight line.

THEOREMS AND EXAMPLES ON BOOK III.

259

Let AB be the given straight line, and K the side of the given
square.

Y' D t

X A M'

M B

It is required to divide the at. line AB at apoint M, so that
the red, AM, MB may he equal to the sq. on K.

Adopting a construction suggested by ii. 14,

describe a semicircle on AB ; and at any point X in AB, or AB
produced, draw XY perp. to AB, and equal to T<.

Through Y draw YZ par^ to AB, to meet the arc of the semicircle
at P.

Then if the perp. PM is drawn to AB, it may be shewn after the
manner of ii. 14, or by iii. 35 that

the rect. AM, MB = the sq. on PM

=the sq. on K.

So that the rectangle AM, MB increases as K increases.

Now if K is less than the radius CD, then YZ will meet the arc
of the semicircle in two points P, P' ; and it follows that AB may be
divided at tivo points, so that the rectangle contained by its segments
may be equal to the square on K. If K increases, the st. line YZ
will recede from AB, and the points of intersection P, P' will con-
tinually approach one another ; until, when K is equal to the radius
CD, the St. line YZ (now in the position Y'Z') will meet the arc in
two coincident points^ that is, will touch the semicircle at D ; and
there will be only one solution of the problem.

If K is greater than CD, the straight line YZ will not meet the
semicircle, and the problem is impossible.

Hence the greatest length that K may have, in order that the
construction may be possible, is the radius CD.

.*. the rect. AM, MB is a maximum, when it is equal to the
square on CD ;

that is, when PM coincides with CD, and consequently when M
is the middle point of AB.

Note. The special feature to be noticed in this problem is that
the maximum is found at the transitional point between two solutions
and no solution ; that is, when the two solutions coincide and be-
come identical.

260 extclid's elements.

The following example illustrates the same point.

2. To find at \ohcU point in a given straight line the angle sub-
tended by the line joining two given points, which are on the same, side
of the given straight line, is a maximum.

Let CD be the given st. line, and A, B the given points on the
same side of CD.

It is required to find at whai point in CD the angle subtended by the
st, line AB is a maximum.

First determine at what point in CD, the st. line AB subtends a
given angle.

This is done as follows: —

On AB describe a segment of a circle containing an angle equal to
the given angle. in. 33.

If the arc of this segment intersects CD, (wo points in CD are
found at which AB subtends the given angle : but if the arc does
not meet CD, wo solution is given.

In accordance with the principles explained above, we expect
that a maximum angle is determined at the limiting position ; that
is, when the arc touches CD, or meets it at two coincident points.

[See page 231.]

This we may prove to be the case.

Describe a circle to pass through A and
B, and to touch the st. line CD. q

[Ex. 21, p. 253.] \

Let P be the point of contact. ^, — ^SP

Then shall the L APB be greater than y' /\\^Q

any other angle subtended by AB at a point / / I jjfjj

in CD on the same side of AB as P. •' / ,]''/-/K\D

For take Q, any other point in CD, on \ / ,-:''' \ •''/

the same side of AB as P ; \ A"''' l//

and join AQ, QB. ^<;- - v>g

Since Q is a point in the tangent other " '
than the point of contact, it must be with-
out the circle ;

.'. either BQ or AQ must meet the arc of the segment APB.
Let BQ meet the arc at K : join AK.

Then the L APB = the L AKB, in the same segment :
but the ext. L AKB is greater than the int. opp. L AQB.

.*. the L APB is greater than AQB.

Similarly the L APB may be shewn to be greater than any other
angle subtended by AB at a point in CD on the same side of AB :
that is, the L APB is the greatest of all such angles. q.e.d.

Note. Two circles may be described to pass through A and B,
and to touch CD, the points of contact being on opposite sides of AB ;

I*'

THEOREMS AND EXAMPLES ON BOOK III. 261

hence two points in CD may be found such that the angle subtended
by AB at each of them is greater than the angle subtended at any
other point in CD oti the same side o/AB.

We add two more examples of considerable importance.

3. In a straight line of indefinite length find a point such that the
sum of its distances from tioo given points, on the same side of the given
line, shall be a minimum.

Let CD be the given st. line of
indefinite length, and A, B the
given points on the same side of
CD.

It is required to find a point P in
CD, awc/t tha^ the sum o/AP, PB is
a minimum.

Draw AF perp. to CD ;
and produce AF to E, making FE
equal to AF.

Join EB, cutting CD at P. E

Join AP, PB.

Then of all lines drawn from A and S to a point in CD,
the sum of AP, PB shall be the least.

For, let Q be any other point in CD.
Join AQ, BQ, EQ.

Now in the A« AFP, EFP,
r AF = EF, Constr.

Because i and FP is common ;

I and the L AFP = the Z- EFP, being rt. angles.
.-. AP=EP. I. 4.

Similarly it may be shewn that
AQ=EQ.

Now in the A EQB, the two sides EQ, QB are together greater
than EB ;

hence, AQ, QB are together greater than EB,
that is, greater than AP, PB.

Similarly the sum of the st. lines drawn from A and B to any
other point in CD may be shewn to be greater than AP, PB.

.*. the sum of AP, PB is a minimum.

Q.E.D.

Note. It follows from the above proof that

the L APF =the Z. EPF i. 4.

= the L BPD. I. 15.

Thus the sum of AP, PB is a minimum, when these lines are
equally inclined to CD.

262 EUCLID'S ELEMENTS.

4. Given two intersecting straight lines AB, AC, and a point P
between them ; shew that of all straight lines which pass through P
and are terminated &y AB, AC, that which is bisected at P cvts off the
triangle of minimum area.

Let EF be the st. line, terminated
by AB, AC, which is bisected at P.

Then the A FAE shall be of mini-
mum area.

y^ /P

For let HK be any other st. line ^ /\ .

passing through p. y/^ /.iM

Through E draw EM par^ to AC.

Then in the A» HPF, MPE,

f the Z- HPF = the Z. MPE, i. 15.

Because \ and the L HFP = the L MEP, I. 29.

I andFP=EP; Hyp,

:. the A HPF = the A MPE. i. 26, Cor.

But the A MPE is less than the A KPE ;
.-. the A HPF is less than the A KPE :
to each add the fig. AHPE ;
then the A FAE is less than the A HAK.

Similarly it may be shewn that the A FAE is less than any other
triangle formed by drawing a st. line through P :

that is, the A FAE is a minimum.

Examples.

1. Two sides of a triangle are given in length ; how must they
be placed in order that the area of the triangle may be a maximum f

2. Of all triangles of given base and area, the isosceles is that
which has the least perimeter.

3. Given the base and vertical angle of a triangle ; construct it
so that its area may be a maximum.

4. Find a point in a given straight line such that the tangents
drawn from it to a given circle contain the greatest angle possible.

5. A straight rod slips between two straight rulers placed at
right angles to one another ; in what position is the triangle inter-
cepted between the rulers and rod a maximum ?

THEOREMS AND EXAMPLES ON BOOK III. 263

6. Divide a given straight line into two parts, so that the sum
of the squares on the segments

(i) may be equal to a given square ;

(ii) may be a minimum.

7. Through a point of intersection of two circles draw a straight
line terminated by the circumferences,

(i) so that it may be of given length ;

(ii) so that it may be a maximum.

8. Two tangents to a circle cut one another at right angles ;
find the point on the intercepted arc such that the sum of the
perpendiculars drawn from it to the tangents may be a minimum.

9. Straight lines are drawn from two given points to meet one
another on the convex circumference of a given circle : prove that
their sum is a minimum when they make equal angles with the
tangent at the point of intersection.

10. Of all triangles of given vertical angle and altitude, that
which is isosceles has the least area.

11. Two straight lines CA, CB of indefinite length are drawn
from the centre of a circle to meet the circumference at A and B ;
then of all tangents that may be drawn to the circle at points on the
arc AB, that whose intercept is bisected at the point of contact cuts
o£f the triangle of minimum area.

12. Given two intersecting tangents to a circle, draw a tancent
to the convex arc so that the triangle formed by it and the given
tangents may be of maximum area.

13. Of all triangles of given base and area, that which is isosceles
has the greatest vertical angle.

14. Find a point on the circumference of a circle at which the
straight line joining two given points (of which both are within, or
both without the circle) subtends the greatest angle.

15. A bridge consists of three arches, whose spans are 49 ft.,
32 ft. and 49 ft. respectively : shew that the point on either bank
of the river at which the middle arch subtends the greatest angle is
63 feet distant from the bridge.

16. From a given point P without a circle whose centre is C,
draw a straight line to cut the circumference at A and B, so that the
triangle ACo may be of maximum area.

17. Shew that the greatest rectangle which can be inscribed in
a circle is a square.

18. A and B are two fixed points w^ithout a circle : find a point
P on the circumference, such that the sum of the squares on Ar, PB
may be a minimum. [See p. 161, Ex. 24.]

264 Euclid's elements.

19. A segment of a circle is described on the chord AB : find a
point C on its arc so that the sum of AC, BC may be a maximum.

20. Of all triangles that can he inscribed in a circle that which has
the greatest perimeter is equilateral.

21. Of all triangles that can he inscribed in a given circle that
which has the greatest area is equilateral.

22. Of all triangles that can he inscribed in a given triangle thai
which has the least perimeter is the triangle formed by joining the feet
of the perpendiculars drawn from the vertices on opposite sides,

23. Of all rectangles of given area, the square has the least
perimeter.

24. Describe the triangle of maximum area, having its angles
equal to those of a given triangle, and its sides passing through three
given points.

VI. HARDER MISCELLANEOUS EXAMPLES.

1. AB is a diameter of a given circle ; and AC, BD, two chords
on the same side of AB, intersect at E : shew that the circle which
passes through D, E, C cuts the given circle orthogonally.

2. Two circles whose centres are C and D intersect at A and B ;
and a straight line PAQ is drawn through A and terminated by the
circumferences : prove that

(i) the angle PBQ=the angle CAD

(ii) the angle BPC = the angle BQD.

3. Two chords AB, CD of a circle whose centre is O intersect at
right angles at P : shew that

(i) PA2 + PB2 + PC2 + PD2 = 4 (radius)2.

(ii) AB2 + CD2 + 40F =8 (radius)^.

4. Two parallel tangents to a circle intercept on any third
tangent a portion which is so divided at its point of contact that the
rectangle contained by its two parts is equal to the square on the
radius.

5. Two equal circles move between two straight lines placed
at right angles, so that each straight line is touched by one circle,
and the two circles touch one another : find the locus of the point
of contact.

6. AB is a given diameter of a circle, and CD is any parallel
chord : if any point X in AB is joined to the extremities of CD,
shew that

XC2+XD2=XAa+XBa.

THEOREMS AND EXAMPLES ON BOOK III. 265

7. PQ is a fixed chord in a circle, and PX, QY any two parallel
chords through P and Q ; shew that XY touches a fixed concentric
circle.

8. Two equal circles intersect at A and B ; and from C, any
point on the circumference of one of them, a perpendicular is drawn
to AB, meeting the other circle at O and O'; shew that either O or
O' is the orthocentre of the triangle ABC. Distinguish between the
two cases.

9. Three equal circles pass through the same point A, and their
other points of intersection are B, C, D : shew that of the four
points A, B, C, D, each is the orthocentre of the triangle formed by
joining the other three.

10. From a given point without a circle draw a straight line
to the concave circumference so as to be bisected by the convex
circumference. When is this problem impossible ?

11. Draw a straight line cutting two concentric circles so that
the chord intercepted by the circumference of the greater circle may
be double of the chord intercepted by the less.

12. ABC is a triangle inscribed in a circle, and A', B', C are the
middle points of the arcs subtended by the sides (remote from the
opposite vertices) : find the relation between the angles of the two
triangles ABC, A'B'C' ; and prove that the pedal triangle of A'B'C
is equiangular to the triangle ABC.

13. The opposite sides of a quadrilateral inscribed in a circle
are produced to meet : shew that the bisectors of the two angles so
formed are perpendicular to one another.

14. K a quadrilateral can have one circle inscribed in it, and
another circumscribed about it ; shew that the straight lines joining
the opposite points of contact of the inscribed circle are perpendicular
to one another.

15. Given the base of a triangle and the sum of the remaining
sides ; find the locus of the foot of the perpendicular from one
extremity of the base on the bisector of the exterior vertical angle.

16. Two circles touch each other at C, and straight lines are
drawn through C at right angles to one another, meeting the circles
at P, P' and Q, Q' respectively : if the straight line which joins the
centres is terminated by the circumferences at A and A', shew that

P'F + Q'Q2=A'A2.

17. Two circles cut one another orthogonally at A and B ; P
is any point on the arc of one circle intercepted by the other, and
PA, PB are produced to meet the circumference of the second circle
at C and D : shew that CD is a diameter.

266 EUCLID'S ELEMENTS.

18. ABC is a triangle, and from any point P perpendiculars
PD, PE, PF are drawn to the sides : if Sj, Og, S^are the centres of
the circles circumscribed about the triangles EPF, FPD, DPE,
shew that the triangle S^SgSg is equiangular to the triangle ABC,
and that the sides of the one are respectively half of the sides of the
other.

19. Two tangents PA, PB are drawn from an external point P
to a given circle, and C is the middle point of the chord of contact
AB ; if XY is any chord through P, shew that AB bisects the angle
XCY.

20. Given the sum of two straight lines and the rectangle con-
tained by them (equal to a given square) : find the lines.

21. Given the sum of the squares on two straight lines and the
rectangle contained by them : find the lines.

22. Given the sum of two straight lines and the sum of the
squares on them : find the lines.

23. Given the difference between two straight lines, and the
rectangle contained by them : find the lines.

24. Given the sum or difference of two straight lines and the
difierence of their squares : find the lines.

25. ABC is a triangle, and the internal and external bisectors
of the angle A meet BC, and BC produced, at P and P' : if O is the
middle point of PP', shew that OA is a tangent to the circle circum-
scribed about the triangle ABC.

26. ABC is a triangle, and from P, any point on the circum-
ference of the circle circumscribed about it, perpendiculars are drawn
to the sides BC, CA, AB meeting the circle again in A', B', C;
prove that

(i) the triangle A'B'C is identically equal to the triangle ABC.

(ii) AA', BB', CC are parallel.

27. Two equal circles intersect at fixed points A and B, and
from any point in AB a perpendicular is drawn to meet the circum-
ferences on the same side of AB at P and Q : shew that PQ is of
constant length.

28. The straight lines which join the vertices of a triangle to
the centre of its circumscribed circle, are perpendicular respectively
to the sides of the pedal triangle.

29. P is any point on the circumference of a circle circumscribed
about a triangle ABC ; and perpendiculars PD, PE are drawn from
P to the sides BC, CA. Find the locus of the centre of the circle
circumscribed about the triangle PDE.

THEOREMS AND EXAMPLES ON BOOK III. 267

30. P is any point on the circumference of a circle circumscribed
about a triangle ABC : shew that the angle between Simson's Line
for the point P and the side BC is equal to the angle between AP
and the diameter of the circumscribed circle through A.

31. Shew that the circles circumscribed about the four triangles
formed by two pairs of intersecting straight lines meet in a point.

32. Shew that the orthocentres of the four triangles formed by
two pairs of intersecting straight lines are collinear.

On the Construction of Triangles.

33. Given the vertical angle, one of the sides containing it, and
the length of the perpendicular from the vertex on the base : con-
struct the triangle.

34. Given the feet of the perpendiculars drawn from the vertices
on the opposite sides : construct the triangle.

35. Given the base, the altitude, and the radius of the circum-
scribed circle : construct the triangle.

36. Given the base, the vertical angle, and the sum of the
squares on the sides containing the vertical angle : construct the
triangle.

37. Given the base, the altitude and the sum of the squares on
the sides containing the vertical angle : construct the triangle.

38. Given the base, the vertical angle, and the difference of the
squares on the sides containing the vertical angle : construct the
triangle.

39. Given the vertical angle, and the lengths of the two medians
drawn from the extremities of the base : construct the triangle.

40. Given the base, the vertical angle, and the difference of the
angles at the base : construct the triangle.

41. Given the base, and the position of the bisector of the
vertical angle : construct the triangle.

42. Given the base, the vertical angle, and the length of the
bisector of the vertical angle : construct the triangle.

43. Given the perpendicular from the vertex on the base, the
bisector of the vertical angle, and the median which bisects the
base : construct the triangle.

44. Given the bisector of the vertical angle, the median bisect-
ing the base, and the difference of the angles at the base : construct
the triangle.

BOOK IV.

Book IV. consists entirely of problems, dealing with
various rectilineal figures in relation to the circles which
pass through their angular points, or are touched by their
sides.

Definitions.

1. A Polygon is a rectilineal figure bounded by more
than four sides.

A Polygon of five sides is called a Pentagon,

six sides „ Hexagon,

sevm sides „ Heptagon,

eight sides „ Octagon,

ten sides „ Decagon,

twelve sides „ Dodecagon,

fifteen sides „ Quindecagon.

2. A Polygon is Eegular when all its sides are equal,
and all its angles are equal.

3. A rectilineal figure is said to be
inscribed in a circle, when all its angular
points are on the circumference of the circle;
and a circle is said to be circumscribed
about a rectilineal figure, when the circum-
ference of the circle passes through all the
angular points of the figure.

4. A circle is said to be inscribed in a
rectilineal figure, when the circumference of
the circle is touched by each side of the figure ;
and a rectilineal figure is said to be circum-
scribed about a circle, when each side of the
figure is a tangent to the circle.

5. A straight line is said to be placed in a circle, when
its extremities are on the circumference of the circle.

BOOK IV. PROP. 1. 269

Proposition 1. Problem.

In a given circle to place a chord eqrml to a given straight
line, which is not greater than the diameter of the circle.

Let ABC be the given circle, and D the given straight
line not greater than the diameter of the circle.

// is required to place in the O ABC a chord equal to D.

Construction. Draw CB, a diameter of the O ABC.
Then if CB = D, the thing required is done.
But if not, CB must be greater than D. Hyp,

From CB cut off CE equal to D : I. 3.

and with centre C, and radius CE, describe the AEF,
cutting the given circle at A.
Join CA.

Then CA shall he the chord required.

Proof. For CA = CE, being radii of the AEF;

and CE = D : Constr,

.-. CA=D.

Q.E.F.

EXERCISES.

1. In a given circle place a chord of given length so as to pass
through a given point (i) without, (ii) within the circle.

When is this problem impossible ?

2. In a given circle place a chord of given length so that it may
be parallel to a given straight line.

270 kuclid's elements.

Proposition 2. Problem.

In a given circle to inscribe a triangle equiangular to a given
triangle.

Let ABC be the given circle, and DEF the given triangle.

It is required to inscribe in the ABC a triangle equiangular to
the A DEF.

Construction, At any point A, on the O*^ of the ABC,
draw the tangent GAH. IIL 17.

At A make the l GAB equal to the l DFE ; I. 23.
and make the z. HAC equal to the z. DEF. I. 23.

Join BC.

Then ABC shall be the triangle required.

Proof. Because GH is a tangent to the ABC,
and from A its point of contact the chord AB is drawn,
.'. the L GAB = the l ACB in the alt. segment : III. 32.
but the L GAB = the L DFE ; Constr,

.-. the z. ACB = the z_ DFE.

Similarly the l HAC = the l ABC, in the alt. segment :
.-. the L ABC = the l DEF. Constr,

Hence the third l BAC = the third l EDF,
for the three angles in each triangle are together equal to
two rt. angles. i. 32.

.*. the A ABC is equiangular to the A DEF, and it is
inscribed in the O ABC.

QE.F.

book iv. prop. 3. 271

Proposition 3. Problem.

About a given circle to circumscribe a triangle equiangular
to a given triangle.

F H

Let ABC be the given circle, and DEF the given triangle.

// is required to circumscribe about the O ABC a triangle equi-
angular to the A DEF.

Construction. Produce EF both ways to G and H.

Find K the centre of the O ABC, III. 1.

and draw any radius KB.
At K make the l BKA equal to the l DEG ; I. 23.
and make the l BKC equal to the l DFH.
Through A, B, C draw LM, MN, NL perp. to KA, KB, KC.

27ien LMN shall be the triangle required.

Proof. Because LM, MN, NL are drawn perp. to radii
at their extremities.

.-. LM, MN, NL are tangents to the circle. III. 16.

And because the four angles of the quadrilateral AKBM
together = four rt. angles ; I. 32. Cor.

and of these, the l" KAM, KBM are rt. angles ; Constr.
.". the L* AKB, AMB together = two rt. angles.

But the /J DEG, DEF together = two rt. angles; I. 13.
.-. the L^ AKB, AMB = the L^ DEG, DEF ;
and of these, the l AKB = the l DEG ; Constr,

.', the z. AMB = the ^ DEF.

Similarly it may be shewn that the l. LNM = the l DFE.
.-. the third l MLN = the third l EDF. I. 32.

.*. the A LMN is equiangular to the A DEF, and it is
circumscribed about the O ABC. q.e.f.

272 EUCLID'S ELEMENTS.

Proposition 4. Problem.
To inscribe a circle in a given triangle

B F C

Let ABC be the given triangle.
It is required to inscribe a circle in the A ABC.

Construction. ^ Bisect the l" ABC, ACB by the st. lines

Bl, CI, which intersect at I. I. 9.

From I draw IE, IF, IG perp. to AB, BC, CA. L 12.

Proof. Then in this A" EIB, FIB,

(the lEB\= the l FBI ; Constr.

and the z. BEI = the l BFI, being rt. angles ;
and Bl is common ;

.-. IE = IF. L 26.

Similarly it may be shewn that IF = IG.
.*. IE, IF, IG are all equal.

With centre I, and radius IE, describe a circle.

This circle must pass through the points E, F, G ;
and it will be inscribed in the A ABC.

For since IE, IF, IG, being equal, are radii of the O EFG ;

and since the l" at E, F, G are rt. angles ; Constr,
,'. the EFG is touched at these points by AB, BC, CA :

m. 16.
.•. the EFG is inscribed in the A ABC.

Q.E.F.

BOOK. IV. PROP. 4.

273

Note. From page 111 it is seen that if Al is joined, then Al
bisects the angle BAu : hence it follows that

The bisectors of the angles of a triangle are concurrent, the point oj
intersection being the centre of the inscribed circle.

The centre of the circle inscribed in a triangle is usually called its
in-centre.

Definition.

A circle which touches one side of a triangle and the
other two sides produced is said to be an escribed circle of
the triangle.

To draw an escribed circle of a given triangle.

Let ABC be the given triangle, of which
the two sides AB, AC are produced to E
and F.

It is required to describe a circle touching
BC, and AB, AC produced.

Bisect the Z-« CBE, BCF by the st.
lines Bli, Clj, which intersect at Ij. i. 9.

From li draw I^G, IjlH, I^K perp. to
AE, BC> AF. I. 12.

Then in the A" liBG, IjBH,

the L liBG = the L liBH, Constr.
and the L liGB = the L I^HB,
being rt. angles ;

also I^B is common ;
.-. liG = liH.
Similarly it may be shewn that IjH = liK ;

.*. I^G, liH, IjK are all equal.

With centre Ij and radius I^G, describe a circle.

This circle must pass through the points G, H, K ;
and it will be an escribed circle of the A ABC.

For since LH, IjG, IjK, being equal, are radii of the O HGK,
and since the angles at H, G, K are rt. angles,
.*. the GHK is touched at these points by BC, and by AB, AC
produced :

.*. the O GHK is an escribed circle of the A ABC. q.e.f.

It is clear that every triangle has three escribed circles.

Note. From page 112 it is seen that if Alj is joined, then Ali
bisects the angle BAO : hence it follows that

The bisectors of two exterior angles of a triangle and the bisector oj
the third angle are concurrent, the point of intersection being the centre
of an escribed circle.

h.s.s* s

Because

274 Euclid's elements.

Proposition 5. Problem.
To circumscribe a circle about a given triangle.

A. ^ A A

Because -

Let ABC be the given triangle.
It is required to circumscribe a circle about the A ABC.

Construction. Draw DS bisecting AB at rt. angles; I. 11.
and draw ES bisecting AC at rt. angles.
Then since AB, AC are neither par\ nor in the same st. line,
.•. DS and ES must meet at some point S.

Join SA ;
and if S be not in BC, join SB, SC.

Proof. Then in the A' ADS, BDS,

AD = BD,
and DS is common to both ;
and the l ADS = the l BDS, being rt. angles ;
.-. SA = SB. L 4.

Similarly it may be shewn that SC = SA.
.*. SA, SB, SC are all equal.

With centre S, and radius SA, describe a circle :
this circle must pass through the points A, B, C, and is
therefore circumscribed about the A ABC. Q.E.F.

It follows that

(i) when the centre of the circumscribed circle falls
within the triangle, each of its angles must be acute, for
each angle is then in a segment greater than a semicircle :

(ii) when the centre falls on one of the sides of the
triangle, the angle opposite to this side must be a right
angle, for it is the angle in a semicircle :

BOOK IV. PROP. 5. 275

(iii) when the centre falls without the triangle, the
angle opposite to the side beyond which the centre falls,
must be obtuse, for it is the angle in a segment less than a
semicircle.

Therefore, conversely, if the given triangle he acute-angled, the
centre of the circmnsaribed circle falls within it : if it he a right-
angled triangle, the centre falls on the hypotenuse : if it he an
ohtuse-angled triangle, the centre falls without the triangle.

Note. From page 111 it is seen that if S is joined to the middle
point of BC, then the joining line is perpendicular to BC.

Hence the perpendiculars draian to the sides of a triangle from
their middle points are concurrent, the point of intersection being the
centre of the circle circumscribed about the triangle.

The centre of the circle circumscribed about a triangle is usually
called its drcuin-centre.

EXERCISES.

On the Inscribed, Circumscribed, and Escribed Circles of a

Triangle.

1. An equilateral triangle is inscribed in a circle, and tangents
are drawn at its vertices, prove that

(i) the resulting figure is an equilateral triangle :

(ii) its area is four times that of the given triangle.

2. Describe a circle to touch two parallel straight lines and a
third straight line which meets them. Shew that two such circles
can be drawn, and that they are equal.

3. Triangles which have equal bases and equal vertical angles
have equal circumscribed circles.

4. I is the centre of the circle inscribed in the triangle ABC, and
Ij is the centre of the circle which touches BC and AB, AC produced :
shew thai A, I, 1^ are collinear.

5. If the inscribed and circumscribed circles of a triajigle are con-
centric, shew thai the triangle is equilateral; and that the diameter of
the circumscribed circle is double that of the inscribed circle.

6. ABC is a triangle, and I, S are the centres of the inscribed
and circumscribed circles ; if A, I, S are collinear, shew that AB = AC

276 EUCLID'S ELEMENTS.

7. The sum of the diameters of the inscribed and circumscribed
circles of a right-angled triangle is equal to the sum of the sides
containing the right angle.

8. If the circle inscribed in a triangle ABC touches the sides at
D, E, F, shew that the triangle DEF is acute-angled ; and express
its angles in terms of the angles at A, B, C.

9. If I is the centre of the circle inscribed in the triangle ABC,
and Ij the centre of the escribed circle which touches BO ; shew
that I, B, li, C are coney clic.

10. In any triangle the difference of two sides is equal to the
difference of the segments into which the third side is divided at
the point of contact of the inscribed circle.

11. In the triangle ABC the bisector of the angle BAC meets
the base at D, and from I the centre of the inscribed circle a per-
pendicular IE is drawn to BC : shew that the angle BID is equal to
the angle CI E.

12. In the triangle ABC, I and S are the centres of the inscribed
and circumscribed circles : shew that IS subtends at A an angle
equal to half the difference of the angles at the base of the triangle.

13. In a triangle ABC, I and S are the centres of the inscribed
and circumscribed circles, and AD is drawn perpendicular to BC :
shew that Al is the bisector of the angle DAS.

14. Shew that the area of a triangle is equal to the rectangle
contained by its semi -perimeter and the radius of the inscribed circle.

15. The diagonals of a quadrilateral ABCD intersect at O : shew
that the centres of the circles circumscribed about the four triangles
AOB, BOC, COD, DOA are at the angular points of a parallelogram.

16. In any triangle ABC, if I is the centre of the inscribed circle,
and if Al is produced to meet the circumscribed circle at O ; shew that
O is the centre of the circle circumscribed about the triangle BIC.

17. Given the base, altitude, and the radius of the circumscribed
circle ; construct the triangle.

18. Describe a circle to intercept equal chords of given length
on three given straight lines.

19. In an equilateral triangle the radii of the circumscribed and
escribed circles are respectively double and treble of the radius of
the inscribed circle.

20. Three circles whose centres are A, B, C touch one another
externally two by two at D, E, F : shew that the inscribed circle of
the triangle ABC is the circumscribed circle of the triangle DEF.

BOOK IV. PROP. 6.

277

Proposition 6. Problem.

To inscribe a square in a given circle,

Let ABCD be the given circle.
// is required to inscribe a square in the ABCD.

CJonstruction. Find E the centre of the circle : ill. 1.
and draw two diameters AC, BD perp. to one another. I. 11.

Join AB, BC, CD, DA.

Then the fig, ABCD shall he the square required.

Proof.

Because

I. Def. 15.

For in the A" BEA, DEA,
BE = DE,
and EA is common ;
and the l BEA = the l DEA, being rt. angles ;

.-. BA = DA. I. 4.

Similarly it may be shewn that CD = DA, and that BC = CD.

.*. the fig. ABCD is equilateral.

And since BD is a diameter of the ABCD,
.*. BAD is a semicircle ;
.*. the L BAD is a rt. angle. III. 31.

Similarly the other angles of the fig. ABCD are rt. angles.

.•. the fig. ABCD is a square ;
and it is inscribed in the given circle.

Q.E.F.

[For Exercises see page 281.]

278

EUCLID'S ELEMENTS.

Proposition 7. Problem.
To circuTnscrihe a square about a given circle.

H C K

Let ABCD be the given circle.
It is required to circumscribe a square about the ABCD.

Construction. Find E the centre of the ABCD : III. 1.
and draw two diameters AC, BD perp. to one another. I. 11.
Through A, B, C, D draw FG, GH, HK, KF perp. to EA, EB,
EC, ED.

Then the Jig. GK shall be the square required.

Proof. Because FG, GH, HK, KF are drawn perp. to
radii at their extremities,

.*. FG, GH, HK, KF are tangents to the circle. III. 16.

And because the z." AEB, EBG are both rt. angles, Constr.

.'. GH is par^ to AC. L 28.

Similarly FK is par^ to AC :
and in like manner GF, BD, HK are par\

Hence the figs. GK, GC, AK, GD, BK, GE are par

.-. GF and HK each = BD ;

also GH and FK each = AC :

but AC = BD ;

.-. GF, FK, KH, HG are all equal :

that is, the fig. GK is equilateral.

And since the fig. GE is a par™,

.-. the L BGA = the L BEA;

but the L BEA is a rt. angle ;

.'. the z_ at G is a rt. angle.

Similarly the z.' at F, K, H are rt. angles.

.*. the fig. GK is a square, and it has been circumscribed

about the ABCD Q.E.F.

,ms

L 34.

Constr.

BOOK IV. PROP. 8.

279

Proposition 8. Problem.
To inscribe a circle in a given square.

Let ABCD be the given square.
It is required to inscribe a circle in the square ABCD.

Construction. Bisect the sides AB, AD at F and E. I. 10.

Through E draw EH par^ to AB or DC : I. 31.

and through F draw FK pai^ to AD or BC, meeting EH at G.

Proof. Now AB = AD, being the sides of a square ;

and their halves are equal Ax. 7.

.-. AF = AE.
But the fig. AG is a par" ; Constr,

.-. AF = GE, and AE = GF ;
.-. GE = GF.

Similarly it may be shewn that GE = GK, and GK = GH :
.-. GF, GE, GK, GH are all equal.

With centre G, and radius GE, describe a circle.
This circle must pass through the points F, E, K, H ;

and it will be touched by BA, AD, DC, CB ; III. 16.
for GF, GE, GK, GH, being equal, are radii ;
and the angles at F, E, K, H are rt. angles. I. 29.
Hence the O FEKH is inscribed in the sq. ABCD.

Q.£.F.

[For Exercises see p. 281.]

280

EUCLID'S ELEMENTS.

Proposition 9. Problem-
To circumscribe a circle ahovi a given square.

Let ABCD be the given square.
It is required to circumscribe a circle about the square ABCD.

Construction. Join AC, BD, intersecting at E.

Proof. Then in the A" BAC, DAC,

I BA = DA, I. Def. 30.

Because \ and AC is common ;

[and BC = DC ; I. Def. 30.

. •. the L. BAC = the z. DAC ; I. 8.

that is, the diagonal AC bisects the l BAD.

Similarly the remaining angles of the square are bisected
by the diagonals AC or BD.

Hence each of the z." EAD, EDA is half a rt. angle ;
.-. the ^ EAD = the l EDA :

.-. EA=ED. • L 6.

Similarly it may be shewn that ED = EC, and EC = EB.
.-. EA, EB, EC, ED are all equal.

With centre E, and radius EA, describe a circle :
this circle must pass through the points A, B, C, D^ and is
therefore circumscribed about the sq. ABCD. Q.E.F.

BOOK IV. PROP. 9. 281

Definition. A rectilineal figure about which a circle
raay be described is said to be CycliCc

EXERCISES on PROPOSITIONS 6-9.

I. If a cirde can he inscribed in a quadrilateral y sheio that the
€um of one pair of opposite sides is equal to the s^im of the other pair.

2. If the sum of one pair of opposite sides of a quadrilateral is
equal to the sum of the other pair, shew thai a circle may he inscrihed
in the figure,

[Bisect two adjacent angles of the figure, and so describe a circle to
touch three of its sides. Then prove indirectly by means of the
last exercise that this circle must also touch the fourth side.]

3. Prove that a rhombus and a square are the only parallelograms
in which a circle can he inscribed.

4. AU cyclic parallelograms are rectangular.

5. The greatest rectangle which can he inscribed in a given cirde
is a square,

6. Circumscribe a rhombus about a given circle.

7. All squares circumscribed about a given circle are equal.

8. The area of a square circumscribed about a circle is double
of the area of the inscribed square.

9. ABCD is a square inscribed in a circle, and P is any point
on the arc AD : shew that the side AD subtends at P an angle three
times as great as that subtended at P by any one of the other sides.

10. Inscribe a square in a given square ABCD, so that one of its
angular points shall be at a given point X in AB.

II. In a given square inscribe the square of minimum area.

12. Describe (i) a circle, (ii) a square about a given rectangle.

13. Inscribe (i) a circle, (ii) a square in a given quadrant.

14. ABCD is a square inscribed in a circle, and P is any point
on the circumference ; shew that the sum of the squares on PA, PB,
PC, PD is double the square on the diameter. [See Ex. 24, p. 161.]

282 EUCLID'S ELEMENTS.

Proposition 10. Problem.

To describe an isosceles triangle having each of the angles at
the base double of the third angle.

Construction. Take any straight line AB.
Divide AB at C, so that the rect. BA, BC = the sq. on AC.

il 11.
With centre A, and radius AB, describe the O BDE ;
and in it place the chord BD equal to AC. IV. 1.

Join DA.

Then ABD shall be the triangle required.

Join CD ;
and about the A ACD circumscribe a circle. IV. 5.

Proof. Now the rect. BA, BC = the sq. on AC Constr,

= the sq. on BD. Constr.
Hence BD is a tangent to the ACD : III. 37.
and from the point of contact D a chord DC is drawn ;
.-. the L BDC = the l CAD in the alt. segment. IIL 32.

To each of these equals add the z. CDA :
then the whole z. BDA = the sum of the z." CAD, CDA.

But the ext. l DCB = the sum of the z." CAD, CDA; I. 32.

.-. the z. DCB = the z_ BDA.
And since AB = AD, being radii of the O BDE,

.• . the L DBA = the z. BDA ; L 5.

.'. the z_ DBC = the l DCB ;

BOOK IV. PROP. 10. 283

.-. DC=DB; I. 6.

that is, DC = CA : Constr.

.-. the z. CAD = the ^ CDA; 1.5.

.-. the sum of the l* CAD, CDA = twice the angle at A.

fixat the L ADB = the sum of the l^ CAD, CDA ; Proved.

,\ each of the l" ABD, ADB = twice the angle at A.

EXERCISES ON PROPOSITION 10.

1. In an isosceles triangle in which each of the angles at the
base is double of the vertical angle, shew that the vertical angle is
one-fifth of two right angles.

2. Divide a right angle iviojive equal parts.

3. Describe an isosceles triangle whose vertical angle shall be
three times either angle at the base. Point out a triangle of this
kind in the figure of Proposition 10,

4. In the figure of Proposition 10, if the two circles intersect at F,
shew that BD = DF.

5. In the figure of Proposition 10, shew that the circle ACD is
equal to the circle circumscribed about the triangle ABD.

6. In the figure of Proposition 10, if the two circles intersect at
F, shew that

(i) BD, DF are sides of a regular decagon inscribed in the
circle EBD.

(ii) AC, CD, DF are sides of a regular pentagon inscribed
in the circle ACD.

7. In the figure of Proposition 10, shew that the centre of the
circle circumscribed about the triangle DBC is the middle point of
the arc CD.

8. In the figure of Proposition 10, if I is the centre of the circle
inscribed in the triangle ABD, and I', S' the centres of the inscribed
and circumscribed circles of the triangle DBC, shew that S1 = S'r.

284 euclid's elements.

Proposition 11. Problem.
To inscribe a regular pentagon in a given circle,

A
F

BiT / \ ^E

Let ABC be a given circle.
It is required' to inscribe a regular pentagon in the ABC.

Construction. Describe an isosceles A FGH, having eaclz
of the angles at G and H double of the angle at F. rv. IG.
In the O ABC inscribe the A ACD equiangular to the
A FGH, IV. 2.

so that each of the l" ACD, ADC is double of the l CAD.

Bisect the z.' ACD, ADC by CE and DB, which meet the
O''* at E and B. L 9.

Join AB, BC, AE, ED.
Then ABCDE shall be the required regular pentagon.

Proof. Since each of the L" ACD, ADC = twice the l CAD ;

and since the z." ACD, ADC are bisected by CE, DB,

.-. the five z." ADB, BDC, CAD, DCE, ECA are all equal.

.-. the five arcs AB, BC, CD, DE, EA are all equal. IIL 26.

.*. the five chords AB, BC, CD, DE, EA are all equal. III. 29.

.'. the pentagon ABCDE is equilateral.

Again the arc AB = the arc DE ; Proved.

to each of these equals add the arc BCD ;
. •. the arc ABCD = the arc BCDE :
hence the angles at the C® which stand upon these
equal arcs are equal ; III. 27.

that is, the z. AED = the z. BAE.

In like manner the remaining angles of the pentagon
may be shewn to be equal ;

.*. the pentagon ABCDE is equiangular.

Hence the pentagon, being both equilateral and equi-
angular, is regular ; and it is inscribed in the O ABC. Q.KF.

book iv. prop 12. 285

Proposition 12. Problem.
To circumscribe a regvlar pentagon about a given cirde.

K C L

Let ABCD be the given circle.
/ is required to circumscribe a regular pentagon about the
ABCD.
Construction.

Inscribe a regular pentagon in the ABCD, IV. 11.
and let A, B, C, D, E be its angular points.
At the points A, B, C, D, E draw GH, HK, KL, LM, MG,
angents to the circle. ill. 17.

Then shall GHKLM be the required regular pentagon.

Find F the centre of the ABCD ; III. 1.

and join FB, FK, FC, FL, FD.

Proof. In the A" BFK, CFK,

(BF = CF, being radii of the circle,
and FK is common ;
and KB = KC. being tangents to the circle from
the same point K; III. 17, Cor.

.-. the z. BFK = the z. CFK, I. 8.

also the l BKF = the l CKF. I. 8, Cor.

Hence the z. BFC = twice the 2. CFK,
and the l. BKC = twice the z. CKF.
Similarly it may be shewn

that the z. CFD = twice the z_ CFL,
and that the l CLD = twice the z_ CLF.
But since the arc BC = the arc CD, IV. 11.

.-. the ^ BFC = the z. CFD ; III. 27.

and the halves of these angles are equal,
that is, the l. CFK = the z_ CFL.

286

Euclid's elements.

Then in the A» CFK, CFL,
f the L CFK = the l CFL, Proved^

Because-! and the l FCK = the l FCL, being rt. angles, in. 18^
[and FC is common ;

.-. CK = CL, L 26.

and the l FKC = the l FLC.

Hence KL is double of KC; similarly HK is double of KB.
And since KC = KB, lU. 17, Cor,

.'. KL=HK.

In the same way it may be shewn that every two con-
secutive sides are equal ;

.*. the pentagon GHKLM is equilateral.

Again, it has been proved that the z. FKC = the l FLC,
and that the ^' HKL, KLM are respectively double of these
angles :

.-. the L HKL = thez. KLM.

In the same way it may be shewn that every two con-
secutive angles of the figure are equal ;

.*. the pentagon GHKLM is equiangular.

.*. the pentagon is regular, and it is circumscribed about
the ABCD. Q.E.F.

Corollary. Similarly it may he proved that if tangents
are drawn at the vertices of any regidar polygon inscribed in a
circle^ they will form another regular polygon of the same species
drcumscnbed about the circle.

[For Exercises see p. 293.]

BOOK IV. PROP. 13.

287

Proposition 13. Problem.
To inscribe a circle in a given regular pentagon.

Let ABODE be the given regular pentagon.
It is required to inscribe a circle within the figure ABODE.

Construction. Bisect two consecutive z_' BOD, ODE by

I. 9.

CF and DF which intersect at F.

Join FB;
and draw FH, FK perp. to BO, OD.

I. 12.

Proof.

Hyp.

Constr.
1.4.

In the A" BOF, DOF,
[ BO = DO,

Because-! and OF is common to both ;
(and the z. BOF = the z. DOF ;
.-. the z. OBF = the z_ CDF.

But the z. CDF is half an angle of the regular pentagon :
.*. also the lCBF is half an angle of the regular pentagon :

that is, FB bisects the l ABO.

So it may be shewn that if FA, FE were joined, these
lines would bisect the l" at A and E.

Again, in the A" FCH, FCK,
the L FCH = the l FCK, Constr.

and the l FHC = the z. FKO, being rt. angles ;
also FO is common ;

.-. FH = FK. I. 26.

Similarly if FG, FM, FL be drawn perp. to BA, AE, ED,
it may be shewn that the five perpendiculars drawn from F
to the sides of the pentagon are all equal.

Because

288 EUCLID'S ELEMENTS.

With centre F, and radius FH, describe a circle ;

this circle must pass through the points H, K, L, M, G ;

and it will be touched at these points by the sides of the
pentagon, for the z." at H, K, L, M, G are rt. z_". Constr,

.', the HKLMG is inscribed in the given pentagon. Q.E.F.

Corollary. The bisectors of the angles of a regular
pentagon meet at a point.

Note. In the same way it may be shewn that the bisectors of the
angles of any regular polygon meet at a point. [See Ex. 1, p. 294.]

[For Exercises on Regular Polygons see p. 293.]

MISCELLANEOUS EXERCISES.

1. Two tangents AB, AC are drawn from an external point A to
a given circle : describe a circle to touch AB, AC and the convex arc
intercepted by them on the given circle.

2. ABC is an isosceles triangle, and from the vertex A a straight
line is drawn to meet the base at D and the circumference of the
circumscribed circle at E : shew that AB is a tangent to the circle
circumscribed about the triangle BDE.

3. An equilateral triangle is inscribed in a given circle : shew
that twice the square on one of its sides is equal to three times the
area of the square inscribed in the same circle.

4. ABC is an isosceles triangle in which each of the angles at B
and C is double of the angle at A ; shew that the square on AB is
equ^l to the rectangle AB, BC with the square on BC.

BOOK rv. PROP. 14. 289

Proposition 14. Problem.
To circumscribe a circle abord a given regular pentagon.

Let ABODE be the given regular pentagon.
It is required to circumscribe a circle about the figure ABODE.

Construction. Bisect the z_" BCD, ODE by OF, DF, inter-
secting at F. I. 9.

Join FB, FA, FE.

Proof. In the A' BCF, DCF,

r BC = DC, Hyp,

Because -j and CF is common to both ;

[and the l BCF = the l DCF ; Constr.

,\ the L CBF = the l CDF. I. 4.

But the L CDF is half an angle of the regular pentagon :

,•. also the l CBF is half an angle of the regular pentagon :

that is, FB bisects the z. ABC.
So it may be shewn that FA, FE bisect the l* at A and E.

Now the z." FCD, FDC are each half an angle of the
given regular pentagon ;

.-. the z- FCD = the z_ FDC, iv. Def. 2.

,'. FC = FD. I. 6.

Similarly it may be shewn that FA, FB, FC, FD, FE are
all equal.

With centre F, and radius FA, describe a circle :
this circle must pass through the points A, B, C, D, E,
and therefore is circumscribed about the pentagon, q.ej.

Note. In the same way a circle may be circumscribed about any
regular polygon.

H.S.E. T

290 Euclid's elements.

Proposition 15. Problem.
To inscribe a regular hexagon in a given circle.

Let ABDF be the given circle.
It is required to inscribe a regular hexagon in the O ABDF.

Construction. Find G the centre of the O ABDF; III. 1.

and draw a diameter AGD.
With centre D, and radius DO, describe the EGCH.
Join CG, EG, and produce them to cut the O** of the
given circle at F and B.

Join AB, BC, CD, DE, EF, FA.
Then ABODE F shall be the required regular hexagon.

Proof. Now GE = GD, being radii of the ACE ;
and DG = DE, being radii of the EHG :
.*. GE, ED, DG are all equal, and the A EGD is equilateral.
Hence the z_ EGD — one-third of two rt. angles. I. 32.
Similarly the l DGC = one-third of two rt. angles.
But the z." EGD, DGC, CGB together = two rt. angles; L 13.

.*. the remaining l CGB = one third of two rt. angles.
.*. the three ^' EGD, DGC, CGB are equal to one another.

And to these angles the vert. opp. z_" BGA, AGF, FGE
are respectively equal :

.-. the z.' EGD, DGC; CGB, BGA, AGF, FGE are all equal;
.-. the arcs ED, DC, CB, BA, AF, FE are all equal : IIL 26.
.-. the chords ED, DC, CB, BA, AF, FE are all equal : in. 29.

.'. the hexagon is equilateral.

Again the arc FA = the arc DE : Proved,

to each of these equals add the arc ABCD ;

then the arc FABCD = the arc ABCDE :

hence the angles at the O** which stand on these equal arcs

are equal.

BOOK IV. PROP. 15. 291

that is, the l FED = the l, AFE. ill, 27.

In like manner the remaining angles of the hexagon
may be shewn to be equal.

.-. the hexagon is equiangular ;

. *. the hexagon ABCDEF is regular, and it is inscribed in the

ABDF. Q.E.F.

Corollary. Tlie side of a regular hexagon inscribed in
a circle is equal to the radius of the circle.

SUMMARY OF THE PROPOSITIONS OF BOOK IV.

The following summary will assist the student in remembering
the sequence of the Propositions of Book IV.

(i) Of the sixteen Propositions of this Book, Props. 1, 10, 15, 16
deal with isolated constructions.

(ii) The remaining twelve Propositions may be divided into
three groups of four each, as follows :

(a) Group 1. Props. 2, 3, 4, 5 deal with triangles and circles.

(b) Group 2. Props. 6, 7, 8, 9 deal with sqiiares and circles.

circles.

(ill) In each group the problem of inscription precedes the cor-
responding problem of circumscription.

Further, each group deals with the inscription and circumscrip-
tion of rectilineal figures first and of circles afterwards.

292 EUCLID'S ELEMENTS.

Proposition 16. Problem.
To inscribe a regular quindecagon in a given circle,

Let ABCD be the given circle.
It is required to inscribe a regular quindecagon in the O ABCD.

Construction.

In the O ABCD inscribe an equilateral triangle, IV. 2.
and let AC be one of its sides.

In the same circle inscribe a regular pentagon, iv. 11.
and let AB be one of its sides.

Proof.
Now of such equal parts as the whole 0°® contains fifteen,

the arc AC, which is one-third of the O**. contains five,

and the arc AB, which is one-tifth of the O**, contains three;

.*. their difference, the arc BC, contains two.

Bisect the arc BC at E : III. 30.

then each of the arcs BE, EC is one-fifteenth of the O**.

.'. if BE, EC be joined, and st. lines equal to them be
placed succt ssively round the circle, a regular quindecagon
will be inscribed in it. Q.E.F.

BOOK IV. EXERCISES ON PROPS. 11—16. 293

EXERCISES ON PROPOSITIONS 11 — 16.

1. Express in terms of a right angle the magnitude of an angle
of the following regular polygons :

(i) a pentagon, (ii) a hexagon, (iii) an octagon,

(iv) a decagon, (v) a quindecagon.

2. Ai?y angle of a regular pentagon is trisected by the straight
lines which join it to the opposite vertices.

3. In a polygon of n sides the straight lines which join any
angular point to the vertices not adjacent to it, divide the angle
into w - 2 equal parts.

4. Shew how to construct on a given straight line

(i) a regular pentagon, (ii) a regular hexagon, (iii) a regular octagon.

5. An equilateral triangle and a regular hexagon are inscribed
in a given circle ; shew that

(i) the area of the triangle is half that of the hexagon ;

(ii) the square on the side of the triangle is three times the
square on the side of the hexagon.

6. ABODE is a regular pentagon, and AC, BE intersect at H *.
shew that

(i) AB=CH = EH.

(ii) AB is a tangent to the circle circumscribed about the
triangle BHC. '

(iii) AC and BE cut one another in medial section.

7. The straight lines which join alternate vertices of a regular
pentagon intersect so as to form another regular pentagon.

8. The straight lines which join alternate vertices of a regular
polygon of n sides, intersect so as to form another regular polygon
of n sides.

If w=6, shew that the area of the resulting hexagon is one-third
of the given hexagon.

9. By means of iv. 16, inscribe in a circle a triangle whose
angles are as the numbers 2, 5, 8.

10. Shew that the area of a regular hexagon inscribed in a circle
is three-fourths of that of the corresponding circumscribed hexagon.

294 EUCLID'S ELEMENTS.

NOTE ON REGULAR POLYGONS.

The following propositions, proved by Euclid for a regular penta-
gon, hold good for all regular polygons.

1. The bisectors of the angles of any regular polygon are con-
current.

Let D, E, A, B, C be consecutive angular
points of a regular polygon of any number of
sides.

Bisect the L* EAB, ABC by AO, BO, which ^'
intersect at O.

Join EO.

It is required to prove thai EO bisects the L DEA.

For in the A" EAO, BAO,
rEA= BA, being sides of a regular polygon ;
Because \ and AO is common ;

t and the L EAO = the L BAO ; G(mstr.

:. the L OEA = the L OBA. i. 4.

But the L OBA is half the L ABO ; Gmistr,

also the L ABC = the L DEA, since the polygon is regular ;
/. the L OEA is half the L DEA :
that is, EO bisects the L DEA.

Similarly if O be joined to the remaining angular points of the
polygon, it may be proved that each joining line bisects the angle
to whose vertex it is drawn.

That is to say, the bisectors of the angles of the polygon meet at
the point O. Q.E.D.

Corollaries. Since the L EAB = the L ABC ; -^^JP-

and since the L* OAB, OBA are respectively half of the L* EAB, ABC ;

.-. the L OAB = the L OBA ;

.-. OA = OB. L 6.

Similarly OE = OA.

Hence the bisectors of the angles of a regular polygon are all equal.

Therefore a circle described with centre O, and radius OA, wiU
he circumscribed about the polygon.

Also it may be shewn, as in Proposition 13, that perpendiculars
drawn from O to the sides of the polygon are all equal.

Therefore a circle described loith centre O, and any o?ie of these
perpendiculars as radius, wiU be inscribed in the polygon.

BOOK IV. NOTE ON REGULAR POLYGONS. 295

2. If a polygon inscribed in a circle is equHateraly it is also
equiangtUar.

Let AB, BC, CD be consecutive sides of an
equilateral polygon inscribed in the ADK.
Tfien shall this polygon he eqtiiangular.
Because the chord AB = the chord DC. Hyp.
.*. the minor arc AB =the minor arc DC. iii. 28.
To each of these equals add the arc AKD :
then the arc BAKD = the arc AKDC ;
.*. the angles at the 0<^, which stand on these
equal arcs, are equal ;

that is, the L BCD = the L ABC. ill. 27.

Similarly the remaining angles of the polygon may be shewn to
be equal :

the polygon is equiangular. Q.E.D.

3. If a polygon inscribed in a circle is equiangular , it is also
equilateral, provided that the number of its sides is odd.

[Observe that Theorems 2 and 3 are only true of polygons
inacribed in a circle.

Fig. I. Fig. 2.

The above figures are sufficient to shew that otherwise a polygon
may be equilateral without being equiangular, Fig. 1 ; or equiangular
without being equilateral. Fig. 2.]

Note. The following extensions of Euclid's constructions for
Regular Polygons should be noticed.

By continual bisection of arcs, we are enabled to divide the
circumference of a circle,

by means of Proposition 6, into 4, 8,16,..., 2. 2",... equal parts ;
by means of Proposition 15, into 3, 6, 12, ... , 3.2",... equal parts ;
by means of Proposition 1 1 , into 5, 10, 20, ... , 5.2**,... equal parts ;
by means of Proposition 16, into 15, 30, 60,..., 15.2",... equal parts.

Hence we can inscribe in a circle a regular polygon the number
of whose sides is included in any one of the formulae 2.2**, 3.2**,
6 . 2", 15 . 2**, n being any positive integer. It has also been shewn
(by Gauss, 1800) that a regular polygon of 2**+l sides may be
inscribed in a circle, provided 2** + 1 is a prime number.

S&6 Euclid's element^*

QUESTIONS FOR REVISION ON BOOK IV.

1. With what diflference of meaning is the word inscribed used
in the following cases ?

(i) a triangle inscribed in a circle ;

(ii) a circle inscribed in a triangle.

2. What is meant by a cyclic figure ? Shew that aU triangles
are cyclic.

What is the condition that a quadrilateral may be cyclic 1

Shew that cyclic parallelograms must be rectangular.

3. Shew that the only regular figures which may be fitted
together so as to form a plane surface are (i) equilateral triangles^
(ii) sqvjareSy (iii) regular hexagons,

4. Employ the first Corollary of I. 32 to shew that in any

regular polygon of n sides each interior angle contains —5; right

angles ?

5. The bisectors of the angles of a regular polygon are concurrent.
State the method q/"proq/* employed in this and similar theorems.

6. Shew that

(i) all squares inscribed in a given circle are equal ; and

(ii) all equilateral triangles circumscribed about a given
circle are equal.

7. How many circles can be described to touch each of three
given straight lines of unlimited length ?

(i) when no two of the lines are parallel ;

(ii) when two only are parallel ;

(iii) when all three are parallel.

8. Prove that the greatest triangle which can be inscribed in a
circle on a diameter as base, is one-fourth of the circumscribed
square.

9. The radius of a given circle is 10 inches : find the length of
a side of

(i) the circumscribed square ; [20 inches.]

(ii) the insc»:ibed square ; \/2 inches.]

(iii) the inscribed equilateral triangle ; [lOVs inches,]

(iv) the circumscribed equilateral triangle ; [20\/3 inches.]

(v) the inscribed regular hexagon. [10 inches.]

Shew also that the areasof these figures are respectively 400,
200, 75^3, 300\/3, and 160\/3 square inches.

THEOREMS AND EXAMPLES ON BOOK IV.

I. ON THE TRIANGLE AND ITS CIRCLES.

1. D, E, F are the points of contact of the inscribed circle of the
triangle ABC, and Dj, Ej, Fj the points of contact of the escribed
circle, which touches BC and the other sides produced: a, b, c denote
the length of the sides BC, CA, AB ; s the semi-perimeter of the
triangle, and r, r^ the radii of the inscribed and escribed circles,

A
Prove the following eqvdlities :

(i) AE =AF =«-a,
BD=BF =«-6,
CD =CE =5-c,

(ii) AEi=AFi =s.

(ui) CDi = CEi=5-6,
BDi=BFi = 5-c.

(iv) CD =BDi and BD = CDi.

(v) EEi=FFi=a.

(vi) The area of th3 A ABC
=:.rs=r^{s-a).

298

EUCLlD^S ELEMENTS.

2. In the tHangle ABC, I U the centre of the inscribed circle^ and
'u l2» 's ^^'^ centres of the escribed circles touching respectively the sides
BC, CA, AB and the other sides prodticed.

» s

« X

Prove the following p7'operties :

(i) The points A, I, 1^ are collinear: so are B, I, \^; and C, I, Ig.
(ii) The points Ig, A, I3 are collinear ; so are \^ B, Ij ; and
li, C, 1 2.

(iii) The triangles BI^C, CIjA, AljB ai'e equiangidar to one
another.

(iv) The triangle Ijlgls is equiangular to the triangle formed by
joining the poinf.fi of contact of the inscribed circle.

(v) Of the four points I, Ij, I2, Is each is the orthocentre of the
triangle whose vertices are the other three.

(vi) The four circles^ each of which passes through three of the
points I, Ij, 1 2, I3, are aXl eqtud.

THEOREMS AND EXAMPLES ON BOOK IV. 299

3. With the notation of page 297, shew that in a triangle ABC,
if the angle at C is a right angle,

r=8-c; ri=s-b; r^^s-a; r^ = 8.

4. With the figure given on page 298, shew that if the circles
whose centres are I, Ij, Ig, I3 touch Bu at D, Dj, Dg, Dg, then

(i) DD2=Dib8=6. (ii) DD3=DiD2=c.

(iii) D^D^=h + c. (iv) DDi=6«-c.

5. SJiew that the orthocentre and vertices of a triangle are the
centres of the inscribed and escribed circles of the pedal triangle.

[See Ex. 20, p. 243.]

6. Given the base and vertical angle of a triangle, find the locus of
the centre of the inscribed circle. [See Ex. 36, p. 246. ]

7. Given the base and vertical angle of a trianglefftnd the locvs of
the centre of the escribed circle which touches the base.

8. Given the ba^e and vertical angle of a triangle, shew that the
centre of the circumscribed circle is fixed.

9. Given the base BC, and the vertical angle A of a triangle,
find the locus of the centre of the escribed circle which touches AC.

10. Given the base, the vertical angle, and the radius of the
inscribed circle ; construct the triangle.

11. Given the base, the vertical angle, and the radius of the
escribed circle, (i) which touches the base, (ii) which touches one
of the sides containing the vertical angle ; construct the triangle.

12. Given the base, the vertical angle, and the point of contact
with the base of the inscribed circle ; construct the triangle.

13. Given the base, the vertical angle, and the point of contact
with the base, or base produced, of an escribed circle ; construct the
triangle.

14. From an external point A two tangents AB, AC are drawn
to a given circle ; and the angle BAC is bisected by a straight line
which meets the circumference in I and li : shew that I is the centre
of the circle inscribed in the triangle ABC, and Ij the centre of one
of the escribed circles.

15. I ts the centre of the circle inscribed in a triangle, and Ij, Ig, Is
the centres of the escribed circles ; shew that llj, llg, II3 are bisected by
the circumference of the circumscribed circle.

16. ABC is a triangle, and 1 2, 1 3 the centres of the escribed
circles which touch AC, and AB respectively : shew that the points
B, C, I3, I3 lie upon a circle whose centre is on the circumference of
the circle circumscribed about ABC.

300 Euclid's elements.

17. With three given points as centres describe three circles
touching one another two by two. How many solutions will
there be ?

18. Two tangents AB, AC are drawn to a given circle from an
external point A ; and in AB, AC two points D and E are taken so
that DE is equal to the sum of DB and EC : shew that DE touches
the circle.

19. Given the perimeter of a triangle, and one angle in magni-
tude and position : shew that the opposite side always touches a
fixed circle.

20. Given the centres of the three escribed circles ; construct
the triangle.

21. Given the centre of the inscribed circle, and the centres of
two escribed circles ; construct the triangle.

22. Given the vertical angle, perimeter, and the length of the
bisector of the vertical angle ; construct the triangle.

23. Given the vertical angle, perimeter, and altitude ; construct
the triangle.

24. Given the vertical angle, perimeter, and radius of the in-
scril>ed circle ; construct the triangle.

25. Given the vertical angle, the radius of the inscribed circle,
and the length of the perpendicular from the vertex to the base ;
construct the triangle.

26. Given the base, the difference of the sides containing the
vertical angle, and the radius of the inscribed circle ; construct the
triangle. [See Ex. 10, p. 276.]

27. Given a vertex, the centre of the circumscribed circle, and
the centre of the inscribed circle, construct the triangle.

28. In a triangle ABC, I is the centre of the inscribed circle ;
shew that the centres of the circles circumscribed about the triangles
BIC, CIA, AIB lie on the circumference of the circle circumscribed
about the given triangle.

29. In a triangle ABC, the inscribed circle touches the base BC
at D ; and r, rj are the radii of the inscribed circle and of the
escribed circle which touches BC : shew that r . ri = BD . DC.

30. ABC is a triangle, D, E, F the points of contact of its
inscribed circle ; and D'E'F' is the pedal triangle of the triangle
DEF : shew that the sides of the triangle D'E'F' are parallel to
those of ABC.

31. In a triangle ABC the inscribed circle touches BC at D.
Shew that the circles inscribed in the triangles ABD, ACD touch
one another.

THEOREMS AND EXAMPLES ON BOOK IV. 301

On the Nine-Points Circle.

32. In any triangle the middle points of the sides, the feet of the
perpendiculars drawn from the vertices to the opposite sides^ and the
middU points of the lines joining the orthocentrt to the vertices are
cancyclic.

In the A ABC, let X, Y, Z be
the middle points of the sides BC,
CA, AB ; let D, E, F be the feet
of the perp^ drawn to these sides
from A, B, C ; let O be the ortho-
centre, and a, fl, 7 the middle
points of OA, OB, OC.

Then shaJl the nine points X, Y, Z,
D, E, F, a, /3, 7 be concyclic.

Join XY, XZ, Xa, Ya, Za.

Now from the A ABO,
since AZ = ZB, and Aa = aO, Hyp.
.*. Za is pari to BO. Ex. 2, p. 104.

And from the A ABC, since BZ = ZA, and BX = XC, Hyp.

:. ZX is par* to AC.
But BO produced makes a rt. angle with AC ; Hyp.
.*. the L XZa is a rt. angle.

Similarly, the L XYa is a rt. angle. i. 29.

.*. the points X, Z, a, Y are concyclic :

that is, a lies on the O*'® of the circle, which passes through X, Y, Z;
and Xa is a diameter of this circle.

Similarly it may be shewn that /3 and 7 lie on the O** of the
circle which passes through X, Y, Z.

Acain, since aDX is a rt. angle. Hyp.

.*. the circle on Xa as diameter passes through D.

Similarly it may be shewn that E and F lie on the circumference
of the same circle.

.*. the points X, Y, Z, D, E, F, a, /3, 7 are concyclic. q.e.d.

From this property the circle which passes through the middle
points of the sides of a triangle is called the Nine-Points Circle ;
many of its properties may be derived from the fact of its being the
circle circumscribed about the pedal triangle.

302 EUCLID'S ELEMBNTS.

33, To prove that

(i) Ike centre of the nijie-pointa circle is the middle point of
the atraiyhl line which joim the orthocentre to the eirctimneribed ceiUre.
K-points circle is half the radius of

In the A ABC, let X, Y, Z be the
middle points of the aides ; D, E, F
the feet of the perp* ; O the ortho-
centre ; 8 and N the centres of the
circumscribed and nine -points circles
respectively.

(i) To prove (hat N is the
middle point o/SO.

It may be shewn that the perp.
to XD from its middle point bisects
80; Ex. U,p. 106.

Similarly the perp. to EY at its
middle point bisects 80 :

. the intersection of the lines which bisect XD and EY at rt. angles

is its centre : iii. 1.

.-. the centre N is the middle point of 80

ie-points circle is half

le-points circle.
(Proved.)

Hence Xa and SO bisect one another at N.
Then from the A'SNX, ONo,
i 8N = 0N,

Becttuse-^ and NX = Na,

land the L SNX = the L ONa ; i. 15.

.. SX = Oa I. 4.

= Ai..
Aad 8X is also par) to Aa,

.-. SA = Xa. I. 33.

But SA is a radius at the circumscribed circle ;
and Xa is a diameter of the nine-points circle ;
.'. the radius of the nine-points circle is half the radiaa of the cir-
cumacribed circle.

THEOREMS AND EXAMPLES ON BOOK IV. 303

(iii) To prove that the centroid is coUinear with points S, N, O.

Join AX and draw ag par^ to SO.
Let AX meet SO at G.
Then from the A AGO, since Aa = aO, and ag is par^ to OG,

.-. A/7=.grG. Ex. 1, p. 104.

And from the A Xa/;, since aN = NX, and NG is par^ to ag,

.-. grG = GX.
/. AG = §ofAX;
.*. G is the centroid of the triangle ABO. Ex. 4, p. 1 13.
That is, the centroid is collinear with the points S, N, O. q.e.d.

34. Oiven the base and vertical angle oj a triangle, find the locus
qfthe centre of the nine-points circle,

35. The nine-points circle of any triangle ABC, whose ortho-
centre is O, is also the nine-points circle of each of the triangles
AOB, BOC, COA.

36. If I, li, Ij, I3 are the centres of the inscribed and escribed
circles of a triangle ABC, then the circle circumscribed about ABC
is the nine-points circle of each of the four triangles formed by
joining three of the points I, Ij, I2, I3.

37. All triangles which have the same orthocentre and the same
circumscribed circle, have also the same nine-points circle.

38. Given the base and vertical angle of a triangle, shew that
one angle and one side of the pedal triangle are constant.

39. Given the base and vertical angle of a triangle, find the
locus of the centre of the circle which passes through the three
escribed centres.

Note. For another important property of the Nine -points Circle
see Miscellaneous Examples on Book VI., Ex. 60.

II. MISCELLANEOUS EXAMPLES.

1. If four circles are described to touch every three sides of a
quadrilateral, shew that their cen^ires are concyclic.

2. If the straight lines which bisect the angles of a rectilineal
figure are concurrent, a circle may be inscribed in the figure.

3. Within a given circle describe three equal circles touching
one another and the given circle.

4. The perpendiculars drawn from the centres of the three
escribed circles of a triangle to the sides which they touch, are
concurrent.

304 EUCLID'S ELEMENTS.

5. Given an anele and the radii of the inscribed and circnm.
scribed circles ; construct the triangle.

6. Given the base, an angle at the base, and the distance be-
tween the centre of the inscribed circle and the centre of the
escribed circle which touches the base ; construct the triangle.

7. In a given circle inscribe a triangle such that two of its sides
may pass through two given points, and the third side be of given
length.

8. In any triangle ABC, I, ij, I2, I3 are the centres of the in-
scribed and escribed circles, and S^, Sg, 83 are the centres of the
circles circumscribed about the triangles BIC, CIA, AIB : shew that
the triangle 8^8283 has its sides parallel to those of the triangle lil^ls*
and is one-fourth of it in area : also that the triangles ABO and
8^8383 have the same circumscribed circle.

9. O is the orthocentre of a triangle ABC : shew that

AO2 + BC2 = BO2 + CA2 = CO2 + AB2 = d2,
where d is the diameter of the circumscribed circle.

10. If from any point within a regular polygon of n sides per-
pendiculars are drawn to the sides, the sum of the perpendiculars is
equal to n times the radius of the inscribed circle.

11. The sum of the perpendiculars drawn from the vertices of a
regular polygon of n slues on any straight line is equal to n times
the perpendicular drawn from the centre of the inscribed circle.

12. The area of a cyclic quadrilateral is independent of the order
in which the sides are placed in the circle.

13. Given the orthocentre, the centre of the nine-points circle,
and the middle point of the base ; construct the triangle.

14. Of all polygons of a given number of sides, which may be
inscribed in a giren circle, that which is regular has the maximum
area and the maximum perimeter.

15. Of all polygons of a given number of sides circumscribed
about a given circle, that which 4s regular has the minimum area
and the minimum perimeter.

16. Given the vertical angle of a triangle in position and magni-
tu^le, and the sum of the sides containing it : nnd the locus of the
centre of the circumscribed circle.

17. P is any point on the circumference of a circle circumscribed
about an equilateral triangle ABC: shew that PA"-i-PB"-|-PC* is
constant.

*^* Book V, is now very rarely read. The subject-matter,
so far as it is introductory to Book FL, is dealt with in a
simpler manner at page 317, in the chapter called ^ Elementary
Principles of Propmiion.^ The student is advised to proceed
at once to that chapter, leaving Book V. in its stricter form
to he studied at a later stage, if it is thought desirable.

BOOK V.

Book V. treats of Ratio and Proportion, and the method adopted
is such as to place these subjects on a basis independent of arith'
metical principles.

The following notation will be employed throughout this section.

Capital letters, A, B^Cf ... will be used to denote the magnitudes
themselves, not any numerical or algebraical measures of them^ and
small letters, m, », p, .«.. will be used to denote whole numbers.
Also it will be assumed that multiplication, in the sense of repeated
addition, can be applied to any magnitude, so that m . -4 or mA
will denote the magnitude A taken m times.

The symbol > will be used for the words greater than, and < for
leas than.

Definitions.

Definition 1. One magnitude is said to be a multiple of another,
when the first contains the second an exa>ct number of times.

Definition 2. One magnitude is said to be a siibmultiple of
another, when the first is contained an exact number of times in the
second.

The following properties of nmltiples will be assumed as self-
evident.

(1) mA >, =, or < mB according as -4 >, =, or < 5 ; and

conversely.

(2) mA+mB+... = m(A + B+...).

(3) 1{A>B, then mA - mB=m{A - B).

(4) mA+nA-\- ... = {m+n+ ...)A,

(5) If wi > n, then mA - nA = (m - ?i) 4 .

(6) m,nA=mn, A=nm . A=n,mA,

306 EUCLID'S ELEMENTS.

Definition 3. The Ratio of one magnitude to another of the same
kind is the relation which the first bears to the second in respect of
qnantuplicity.

The ratio of ^ to 5 is denoted thus, A: B; and A is called the
antecedent, B the consequent of the ratio.

The term quantuplirity denotes the capacity of the first magnitude
to contain the second with or without remainder.

If Jbhe magnitudes are commensurable, their quantuplicity may be
expressed numerically by observing what multiples of the two
magnitudes are equal to one another.

Thus if A=maf and B=na, it follows that nA=^mB, In this

case A = — By and the quantuplicity of A with respect to B is the

arithmetical fraction —

But if the magnitudes are incommensurable, no multiple of the
first can be equal to any multiple of the second, and therefore the
quantuplicity of one with respect to the other cannot exactly be
expressed numerically : in this case it is determined by examining
how the multiples of one magnitude are distributed among the
multiples of the other.

Thus, let all the multiples of A be formed, the scale extending ad
infinitum ; also let all the multiples of B be formed and placed in
their proper order of magnitude among the multiples of A, This
forms the relative scale of the two magnitudes, and the quantuplicity
of A with respect to B is estimated by examining how the multiples
of A are distributed among' those of B in their relative scale.

In other words, the ratio oi A to B is known, if for all integral
values of m we know the multiples nB and {n + l)B between which
mA lies.

In the case of two given magnitudes A and B, the relative scale
of multiples is definite, and is different from that oi A to G, if C
differs from B by any magnitude however small.

For let D be the difference between B and C ; then however
small D may be. it will be possible to find a number m such that
mD>A. In this case, mB and mC would differ by a magnitude
greater than A , and therefore could not lie between the same two
multiples of /I ; so that after a certain point the relative scale of A
and B would differ from that of A and C.

Definition 4. Magnitudes are said to have a ratio to one another,
when the l^ss can be multiplied so as to exceed the other.

BOOK V. DEFINITIONS. 307

Definition 5. The ratio of one magnitude to another is equal
to that of a third magnitude to a fourth, when if any equimultiples
whatever of the antecedents of the ratios are taken, and also any
equimultiples whatever of the consequents, the multiple of one
antecedent is greater than, equal to, or less than that of its con-
sequent, according as the multiple of the other antecedent is greater
than, equal to, or less than that of its consequent.

Thus the ratio -4 to 5 is equal to that of (7 to Z) when
mC >, =, or < nZ) according as mA >, =, or < w^, whatever whole
numbers m and n may be.

Again, let m be any whole number whatever, and n another
whole number determined in such a way that either mA is equal to
nB, or mA lies between uB and (n-\-l)B ; then the definition asserts
that the ratio of A to B is equal to that of C to D if mC=nD when
'mA=nB; or if mC lies between nD and {n+\)D when mA lies
between nB and {n + \)B.

In other words, the ratio of A to B is equal to that of G to D
when the multiples of A are distributed among those of ^ in the
same manner as the multiples of G are distributed among those of D.

When the ratio of ^ to 5 is equal to that of G to D the four
magnitudes are called proportionals. This is expressed by saying
^^ A is to B as G 18 to D" and the proportion is written

A: B::G: D, or A : B = G : D.

A and D are called the extremes, B and G the means ; also D is
said to be a fonrtli proportional to ^, B, and G.

Definition 6. Two terms in a proportion are said to be homo-
logous when they are both antecedents, or both consequents of the
ratios.

Definition 7. The ratio of one magnitude to another is greater
than that of a third magnitude to a fourth, when it is possible to find
equimultiples of the antecedents and equimultiples of the consequents
such that while the multiple of the antecedent of the first ratio
is greater than, or equal to, that of its consequent, the multiple of
the antecedent of the second is not greater, or is less, than that
of its consequent.

This definition asserts that if whole numbers, m and n can be
found such that while mA is greater than nB, mG is not greater
than nD, or while mA =nB, mG is less than wZ), then the ratio of
yl to jS is greater than that of G to D.

If A is equal to B, the ratio of ^ to ^ is called a ratio of
equality.

If ^ is greater than jB, the ratio of ^ to 5 is called a ratio of
greater inequality.

If A is less than B, the ratio of ^ to B is called a ratio of less
inequality.

308 EUCLID'S ELEMENTS.

Definition 8. Two ratios are said to be reciprocal when the
antecedent and consequent of one are the consequent and antecedent
of the other respectively ; thus B : A ia the reciprocal of ^ : ^.

Definition 9. Three magnitudes of the same kind are said to
be proportionals, when the ratio of the first to the second is equal
to that of the second to the third.

Thus Ay Bf G are proportionals if

A: B::B:C.

B is called a mean proportional to A and C, and G is called a
third proportional to A and B.

Definition 10. Three or more magnitudes are said to be in
continued proportion when the ratio of the first to the second is
equal to that of the second to the third, and the ratio of the second
to the third is equal to that of the third to the fourth, and so on.

Definition 11. When there are any number of magnitudes of the
same kind, the first is said to have to the last the raUo compounded
of the ratios of the first to the second, of the second to the third,
and so on up to the ratio of the last but one to the last magnitude.

For example, ii Ay B, G, D, E he magnitudes of the same kind,
-4 : ^ is the ratio compounded of the ratios A : By B : G^ G : D, and
D:E.

This is sometimes expressed by the following notation :

CA :B

A :^=-

B :G

G :D

\.D'.E.

Definition 12. If there are any number of ratios, and a set of
magnitudes is taken such that the ratio of the first to the second is
equal to the first ratio, and the ratio of the second to the third is
equal to the second ratio, and so on, then the first of the set of
magnitudes is said to have to the last the ratio compounded of the
given ratios.

Thus, HA'.ByGiDyEiFhQ given ratios, and if P, Q, RyShe
magnitudes taken so that

then

fiOOK V. I»ftOtS. 1 AND 2. 309

t)efinit%(m 13. When three magnitudes are proportionals, the
first is said to have to the third the duplicate ratio of that which
it has to the second.

Thus if Ai B :'.B .0,

then A is said to have to G the duplicate ratio of that which it has
to B.

Since A:G=\^^\^^

it is clear that the ratio compounded of two equal ratios is the
duplicate ratio of either of them.

Definition 14. When four magnitudes are in continued }yroportion,
the first is said to have to the fourth the triplicate ratio of that
which it has to the second.

It may he shewn as above that the ratio compounded of three
equal ratios is the triplicate ratio of any one of them.

PROPOSITIONS.

Ohs. Of the propositions of Book V., which, it may be noticed
are all theorems, we here give only the more important.

Proposition 1.

HcUios which are equal to tlte same ratio are equal to one another.

Leti4 :^::P:Q,andalso(7:i>::P:Q; thenshall^ :B::G:D.

For it is evident that two scales or arrangements of multiples
which agree in every respect with a third scale, will agree with one
another.

Proposition 2.

If two ratios are equaX^ the antecedent of the second is greater than,
equal to, or less than its consequent according an the antecedent of the
first is greater than, equal to, or less than its conseqiient.

Let A : B :: G : D,

then G>, =, or <D,

according as ^ >, = , or < ^.

This follows at once from Def. 5, by taking m and n each equal
to unity.

310 Euclid's elements.

Proposition 3.

Invertendo or Inversely. If two ratios are equaly their reciprocal
ratios are equal.

Let A : B ::C : Dy

theu shall B : A:: D: C.

For, by hypothesis, the multiples of A are distributed among
those of B in the same manner as the multiples of C are among
those of D.

therefore also, the multiples of B are distributed among those of A
in the same manner as the multiples of D are among those of C.

That is, B :A :: D:C.

Note. This proposition is sometimes enunciated thus :

If four magnitudes are proportionals j they are also proportionals
when taken inversely.

Proposition 4.

Equal maqnitudes have the same ratio to the same magnitude; and
the same magnitude has the same ratio to equal magnitudes.

Let A, By Che three magnitudes of the same kind, and let A be
equal to B ;

then shall A : C :: B : C

and G : A r.G : B.

Since A = B, their multiples are identical and therefore are dis-
tributed in the same way among the multiples of G.

.-. A : G ::B :Gy Def 5.

.*. also, invertendo, G : A :: G : B. v. 3.

Proposition 5.

Of two unequal magnitudes, the greater has a greater ratio to a
third magnitude than the less hus ; and the same magnitude ha^ a
greater ratio to the less of two magnitudes than it has to the greater.

First, let ^ be > ^ ;

then shall ^ : C be > ^ : C.

Since A > B, it will be possible to find m such that mA exceeds
mB by a magnitude greater than C ;

hence if mA lies between nG and (w + \)G0mB < nC :

and if mA =nG, then mB < nG ;

:. A:G>B:G, Def, 7.

BOOK V. PROPS. 6 AND 7. 311

Secondly i let ^ be < -4 ;

then shall G i Bhe>C: A.

For taking m and n as before,

nC > mBy while nC is not > mA ;

.-. G:B>C:A, 2>e/. 7.

Proposition 6»

Magnitiidea which have the same ratio to the same magnitude are
eqiud to one another ; and those to which the same magnitude has the
same ratio are equal to one another.

First, let A: Ci: B:G; •

then shall A = B.

For if A>B, then A : G > B : G,
and if 5 > ^, then B: G>A'.G, v. 5.

which contradict the hypothesis ;

.-. A=B.

Secondly, let G : A :: G : B ;

then shall A=B.

Because G : A : : G : B,
.'. invertendOy A : G : : B : G, V. 3.

.-. A -By

by the first part of the proof.

Proposition 7.

T?uit mjognitvde which has a greater ratio than another has to the
wime magnitude is the greater of the two ; and that magnitude to
which the sam>e has a greater ratio than it has to another magnitude is
the less of the two.

First, let ^ : C7 be > ^ : C;

then shall -4 be > ^.

For \iA=B, then A:G::B:G, v. 4.

which is contrary to the hypothesis.

And if ^ < ^. then A: G<B :G; v. 5.

which is contrary to the hypothesis ;

.-. A>B.

SI 2 EUCLID^S ELEMENTS.

Secondly f let C : ^ be > C : 5 ;

then shall Ahe< B.

For iiA = B, then G .AixGiB, y. 4

which is contrary to the hypothesis.

And \iA>B, then C \ A<C '.B\ v. 5.

which is contrary to the hypothesis ;

/. A<B.

Proposition 8.

Magnitvdes have the same ratio to one another which their equi-
mvltiples have.

Let A^ Bh& two magnitudes ;

then shall A -. B •.-. mA : mB,

If p, q be any two whole numbers,

then m.pA >, = , or < w . qB

according as pA >, = , or < qB.

But m.pA=p.mA, and m. qB=q.mB;

.'. p . mA >, = , or < g . mB

according as pA >, =, or < qB ;

.*. A : B :: mA : mB. Def, 5.

Cor. Let ^ : ^ : : a : D.

Then since A \ B \\ mA : mB,

and G : D 11 nG : ?iZ> ;

/. mA : m^ : : nG : ?ii>. V. 1.

Proposition 9.

If two ratios are equal, and any equimtdtiples of the antecedents
and also of the conseqiients are taken, the multiple of the ,first ante-
cedent has to that of its consequent the same ratio ow the mvltiple of
the other antecedent has to that of its consequent.

LetA:B::G:D;

then shall mA : nB : : mG : nD,

Let p, g be any two whole numbers ;

then because A : B :: G : D,

pm . (7 >, = , or < gn . Z>

according as pm . ^ >, =, or <qn.B, Def. 5.

that is, p . mG >, = , or < g . nD,

according as p . mA >, = , or < g . nB ;

:. mA :nB :: mG : nD. D^. 5.

BOOK V. 1>R01>S. 10 AND 11. S13

Proposition 10.

If four magnittuies of the same kind are proportionals, the first is
greater than, equal to, or less than the third, according as the second
is greater than, equal to, or less than the fourth.

Let A, B, G, Dhe four magnitudes of the same kind such that

A :B::C:D;
then -4 >, =, or <G
according as 5 >, = , or < Z).
JiB>D, then A: B<A:D', V. 5

hntAiB'.iGiD
.-. G:D<A'.D
/. A:D>G: D

.'. A>G, V. 7.

Similarly it may be shewn that

if ^ < D, then A<G,
and if B = D, then A = C*

pEOPOsmoN 11.

Altemando or Alternately. If four magnitudes of the same kind
are proportionals, they are also proportionals when taken alternately.

Let A, B, G, Dhe four magnitudes of the same kind such that

A:B ::C : D;
then shall A: G :: B: D.

Because A : B :: mA : mB, v. 8.

and G : D :: nG : nD ;
.*. mA : mB : : nG : nD. v. 1.

.*. mA >, =, or <nG
according as mB >, =, or < nD. v. 10.

And m and n are any whole numbers ;

.-. A:G :: B iD. Def. 6.

314 EUCLID'S ELEMENTS.

Proposition 12.

Addendo. If any number of magnitiuies of the same kind art
proportiftncdst as one of the antecedents is to its conseqiientj so is tfie
sum of the antecedents to the sum of the consequents.

Let A, B, G, D, E, F, ... be magnitudes of the same kind such
that

A : B :: G : D :: E : F:: ;

then shall A : B :: A+G + E+... : B + D + F+...,

Because A : B :: G : D :: E : F :: ... ,

.'. according as mA>, =, or <nB,

so is mG >, = , or < nD,

and mE>, =, or <.nF,

.'. 6o is mA+mG+mE-{-...>, =, or <nB+nD+nF+ ...
or m[A+G+E+...)>, =, or KniB+D+F-h ...);
and m and n are any whole numbers ;

.-. A : B ::A+G+E+... :B + D+F+..., Def 5,

Proposition 13.

Componendo. If four magnitudes are proportionals^ the sum of
the first and second is to the second as the s^im of the third and fourth
is to the fourth.

Ijet A : B ::G : D;

then shall A + B : B :: G+D : D.

If m be any whole number, it is possible to find another number
n such that mA =nB, or lies between nB and {n+l)B,

.'. mA ■i-mB=mB +nBi or lieahetween mB + nB eaid mB + {n + l)B.

But mA+mB=m{A+B)y and mB+nB={m + 7i)B ;

.'. m(A+B)={m+n)Bf or lies between {m + n)B and {m+n + l)B.

Also because A : B : : G : B,

.'. mG=nD, or lies between nD and (n + l)D; Def 5.

/. w((7+i>) = (m + ?i)Z) or lies between {m-^n)D and (m+w+l)i>;

that is, the multiples oi G+ D are distributed among those of i> in
the same way as the multiples oi A+B among those of B ;

.-. A+B : B :: G+D: D.

Dividendo. In the same way it may be proved that

A-B : B :: G-D : 2>,

or B-A :B ::D-G : A
according as ^ is > or < jB.

BOOK V. PROP. 14.

315

Proposition 14.

JBquali. If there are tivo sets of magnitudes^ such that the
first is to the second of the first set as the frst to the second of the
othf*r setf and the second to the third of the f rut set cw the second to the
third of the other, and so on to the la^t magnitude : then the first is to
the last of the first set cw the first to the last of the other.

First, let there be three magnitudes A, B, G of one set, and
three, P, Q, i?, of another set,

and let A : B

a,nd B :G

then shall A : G

Because A : B

.'. mA : mB

and because B : G

:. mB : nG

.*. , invertendo, nG : mB

Q :

P'.

:Q,

mP : mQ ;

.B,

mQ- 1 nB,

uB : mQ,

V. 8, Got.

V. 9.
V. 3.

Now, if mA > uGy

then mA : mB > nG : mB ;
.*. mP : mQ > nR : mQ,
and .*. mP > nR.

Similarly 7wP= or <ni? according as mA — or <,nG.

.'. A : G :: P : R.

V. 5.
V. 7.

Def 5.

Secondly, let there be any number of magnitudes. A, B, C, ...
L, M, of one set, and the same number P, Q, R, ... Y, Z, of another
set, such that

A I B I'.P :Q,

B : G ::Q :R,

',M:'.

then shall A

:M::

For A

: G :

:^,

and G

:D::

:5;

.'. by the first cafie A

:Di,

'^B,

Eknd so on, imtil finally A

:M =

,Z.

C0BU)TJ<AKT. If A

:B :

: P

'Q.

and B

: G :

:P:

then A

:G :

:«.

Proved.
Hyp.

316 Euclid's elements.

Proposition 15.

U A :B ::X :Y,

and G : B :: Z : Y;

then shall A^G : B :: X + Z: Y.

For since G : B :: Z : Y,

Hyp.

.*., tnvertendoj

B'.G'.:YiZ.

V. 3.

Also A '. B :.X '. Y,

:. , ex cequcUif

A :G::X : Z,

V. 14.

.'. , componendoy

A + G: G ::X + Z:Z,

V. 13.

Again, G : B :: Z : Y,

Hyp.

.'. , ex ceqtialif

A + G:B::X + Z:Y. -

V. 14.

Proposition 16.

If two ratios are equal, their duplicate ratios are eqvjal.

Let A :B ::G : D;
then shall the duplicate ratio of -4 to 5 be equal to that of C' to D.

Let X be a third proportional to A and B, and Y a third pro-
portional to G and D,

so that A : B :: B : X, and G : D :: D :Y;

then because A : B :: G : D^

:. B :X ::D:Y;

.'. , ex osquali, A : X :: G : Y»

But A : X and G : Y are respectively the duplicate ratios of

A : B ernd G : D, Def. \%

.'. the duplicate ratio of A : ^=that oi G ; D.

Note. The converse of this theorem may be readily proved;
namely,

If the duplicates of two ratios are equal, the ratios ihemaelvts
are equal.

ELEMENTARY PRINCIPLES OF PROPORTION.

INTRODUCTION TO BOOK VI.

1. The first four books of Euclid deal with the absolute
equality or inequality of geometrical magnitudes. In
Book VI. such magnitudes are compared by considering
their ratio or relative greatness.

2. The meaning of the words ratio and proportion in
their simplest arithmetical sense may be given as follows :

(i) The ratio of one number to another is the multiple m'
fraction which the first is of the second.

(ii) Four numbers are in proportion when the ratio of the
first to the second is equal to the ratio of the third to the fourth

3. These definitions are however not strictly applic-
able to the purposes of Pure Geometry, for the following
reasons :

(i) Pure Geometry deals only with magnitudes as
represented by diagrams, without measuring them in terms
of a common unit : in other words, it makes no use of
number for the purpose of comparing magnitudes.

(ii) It commonly happens that Geometrical magnitudes
of the same kind are incommensurable, that is, they are such
that it is impossible to express them exactly in terms of some
common unit. Nevertheless it is always possible to express
the arithmetical ratio of two such magnitudes within any
required degree of accuracy. [See Note, p. 131 : also Hall
and Knight's Elementary Algebra, Art. 289.]

318 EUCLID*S ELEMENTS.

4. Accordingly, the object of Euclid's Fifth Book is to
establish the Theory of Proportion on a basis independent
of number. But as Book V. is now very rarely read, we
propose here merely to illustrate algebraically such principles
of proportion as are required before proceeding to Book VI.
The strict treatment of the subject given in Book V. may
be studied at a later stage, if it is thought desirable.

Oha. In what follows the symbol > will be used for the words
greater tharii and < for less than.

5. The following definitions are selected from Book V.

Definition 1. One magnitude is said to be a multiple of
another, when the first contains the second an exact number
of times.

Thus ma is a multiple of a, if m is any whole number.

Definition 2. One magnitude is said to be a submultiple
of another, when the first is contained in the second an
exact number of times.

Thus -^ is a submultiple of a, if m is any whole number.
m

Definition 3. The ratio of one magnitude to another of
the same kind is the relation which the first bears to the
second in regard to quantity ; this is measured by the
fraction which the first is of the second.

Thus if two such magnitudes contain a and h units respectively,
the ratio of the first to the second is expressed by the fraction -.

The ratio of a to J is generally denoted thus, a :b ; and
a is called the antecedent and b the consequent of the ratio.

The two magnitudes compared in a ratio must be of the aame
kind ; for example, both must be lines, or both angles, or both
areas. It is clearly impossible to compare the length of a straight
line with a magnitude of a different kind, such as the area of a
triangle.

INTRODUCTION TO BOOK VI. DEFINITIONS. 319

Definition 5. Four quantities are in proportion, when
the ratio of the first to the second is equal to the ratio of the
third to the fourth.

When tbe ratio of a to 6 is equal to that of x to y, the four
magnitudes are called proportionals. This is expressed by saying
** a is to b as X is to t/j" and the proportion is written

a : h :: X : y ;

or a : b = X : y.

Here a and y are called the extremes, and b and x the means.

(i) Algebraical Test of Proportion. The ratios a : b
and x:y may be expressed algebraically by the fractions

- and -; thus the four magnitudes a, b, x, y are in pro-

portion if a^ic

b'y'

(ii) Geometrical Test of Proportion. The ratio of one
magnitude to another is equal to that of a third magnitude
to a fourth, when if any equimultiples whatever of the ante-
cedents of the ratios are taken, and also any equimultiples
whatever of the consequents, the multiple of one antecedent
is greater than, equal to, or less than that of its consequent,
according as the multiple of the other antecedent is greater
than, equal to, or less than that of its consequent.

Thus the ratio of a to 5 is equal to that of x to y, that
is to say,

a, 5, ic, y are in proportion,

if mx >, =, or < ny,

according as ?na >, = , or < nb,

whatever whole numbers m and n may be.

Note. The Algebraical and Geometrical Tests of Proportion,
though differing widely in method, really determine the same
property ; for each may be deduced from the other. This is fully
explained on the following page.

320 Euclid's elements.

COMPARISON BETWEEN THE ALGEBRAICAL AND GEOMETRICAL

TESTS OF PROPORTION.

(i) 7/" a, b, X, y satisfy the Algebraical test of proportion, to shew
that they also satisfy the geometrical test.

By hypothesis ?=- ;

6 y

and, multiplying both sides by — , where m and n are any whole

numbers, we obtain ^= — >

no ny

thus these fractions are hoth improper^ or both proper , or both equal
to unity ;

hence wa;>, =, or <.7iyj according as 7/ia >, =, or <.nb, which
is the Geometrical test of proportion.

(ii) If a, b, X, y satisfy the Geometrical test of proportion, to
sheio that they also satisfy the Algebraical test.

By hypothesis mx>y =, or <ny, according as wa>, =, or <.nb,
it is required to prove that

a_x

b~y

ft C^

If - is not equal to -, one of them must be the greater.
by

ft ^ tL

Suppose T > - ; then it will be possible to find some fraction —
by m

which lies between them, n and m being positive integers.

a 71

Hence -r>— » (1)

and -<— (2)

y m

From (1), ma>nb;

from (2), mx<ny;

and these contradict the hypothesis.

a OC €L X

Therefore - and - are not unequal ; that is ■=-=-.
by by

Definition 6. Two terms in a proportion are said to be
homologoiis, when they are both antecedents or both consegverUs
of the ratios.

Thus if a: b :: X : y,

(I apd X are homologous ; also b and y are homologous^,

INTRODUCTION TO BOOK VI. DEFINITIONS. 321

Definition 8. Two ratios are said to be reciprocal, when
the antecedent and consequent of one are respectively the
consequent and antecedent of the other.

Thus 6 : a is the reciprocal of a : 6.

Definition 9 Three magnitudes of the same kind are
said to be proportionals, when the ratio of the first to the
second is equal to that of the second to the third.

Thus a, hy c are proportionals if

a ih ::h I c.

Here h is called a mean proportional to a and c ; and c is called
a third proportional to a and h.

When /oz^r magnitudes are in proportion, namely when

a '.h w c\ df

then d is called a fourth proportional to a, 6, and c.

Definition 10. A series of magnitudes of the same kind
are said to be in continued proportion, when the ratios of
the ^rs^ to the second, of the second to the thirds of the third
to the fourth, and so on, are all equal.

Thus a, 6, c, d, e are in continued proportion, if

a\h = h '. c = c: d = d:e;

. 1 . • .. Ob h c d

that IS, if - = -=- = -.

h c d e

Definition 11. When there are any number of magni-
tudes of the same kind, the first is said to have to the last
the ratio compounded of the ratios of the first to the
second^ of the second to the third, and so on up to the
ratio of the last but one to the last magnitude.

Thus if a, 6, c, d, e are magnitudes of the same kind, then a : e
is the ratio compounded of the ratios

a : 6, b : c, c : d^ die.

Note. Algebra defines the ratio compounded of given ratios
as that formed by mtUtiplying together the fractions ^ whidi
represent the given ratios. In the above illustration it will

be seen that on multiplying together the ratios -, -, -, - we

o c Cu e

obtain the ratio -.

H.S.E. X

322 Euclid's elements.

Definition 13. When three magnitudes are propor-
tionals, the first is said to have to the third the duplicate
ratio of that which it has to the second.

Thus if a : h :: h : Cy

then a : c is said to be the duplicate of the ratio a : h.

Note. In Algebra the duplicate of the ratio a : 6 is defined as
the ratio of a^ to b^.

It is easy to show that the two definitions are identical.

For if a:h :; h : c,

then ? = ^.

Now a a h _a a _a^,

that is, a : c : : a? : h\

6. The following theorems from Book V. are here
proved algebraically. Eeference is made to them in
Book VI. under certain technical names.

Theorem 1. By Equal Ratios. Ratios which are equal to
the same ratio are equal to one another.

That is, if a:b =x:y, and c:d = x:y ;

then shall a :b = c : d.

For, by hypothesis, r = -> and ^ = - :

by d y

a _c

b"^

or a:b = c :d.

hence

Theorem 3. Invertendo, or Inversely. If four magni-
tudes are proportioTialSy they are also proportionals taken invei^seiy.

That is, if a:b=x:y,

then shall b'.a — y.x.

Since, by hypothesis, ? = -, it follows that - = ^;

by ax

or bia = y \x.

INTRODUCTION TO BOOK VI. THEOREMS. 323

Theorem 11. Altemando, or Alternately. If four
magnitudes of the same hind are p'oportionals, they are also
proportumals when taken alternately.

That is, if a:b==x:yy

then shall aix = h :y,

a {R
For, by hypothesis, T = "•

Multipljring both sides by -,

we have -.- = -.-:

X y X

that IS, - = -,

X y

or a:x = b :y.

Note. In this theorem the hypothesis requires that a and b shall
be of the same kind, also that x and y shall be of the same kind ;
while the conclxtsion requires that a and x shall be of the same kind,
and also h and y of the same kind.

Theorem 12. Addendo. In a series of equal ratios (the
magnitudes being all of the same kind), as any antecedent is to
its consequmt so is the sum of the antecedents to the sum of the
cimsequents.

That is, if a:x = b :y = c:z=: ... ;
then shaU a:x = a + b + c+ ... :x + y + z+

Let each of the equal ratios -, -, - ... be equal to k,

X y z

Then a = &r, b^ky, c = kzy . . . ;

.*., by addition,

a + b + c+ ...=k(x + y + z+ ...);
a + b + c+ ... _j__a
x + y + z+ ...~~ ~ X
or a:aj=a + 6 + c+.,. ;a; + y + «+....

324 EUCLID'S ELEMENTS.

Theorem 1 3. Componendo. If fawr magnitudes are pro-
pGrtionals, the strni of the first and second is to the second as
the sum of the third and fourth is to the fourth.

That is, if a:b = x:y;

then shall a + b:b = x+y :y.

For, by hypothesis, t = ~ ^

, a . X . a+h x+y

by' by

that is, a + b :b = x + y :y.

Dividendo. Similarly it may be shewn that a-b ib^x-y : y.

Theorem 14. Ex .£quali. If there are three magnitudes
ay b, c of one set, and three magnitudes x, y, z of a/no&er set ;
and if these are so related that

a:b=^x lyA

and b:c = y:z,)

then shall a: c = x:z.

For, by hypothesis, r = -> ^ind -=- ;

.•., by multiplication,
that is.

by c z

a b_x y^

b c y z

a _x

or a:c = x\z.

Theorem 15. If two proportions have the same consequefxtSy

that is, if a: b^xiyA

and c:b = z:yj

then shall a + c :b = x + z:y.

For, by hypothesis, t = -» and t = - j

.-., by addition,

by by

a-\-c x + z

b y '
or a + c:b==x+z:y.

BOOK VI.

Definitions.

1. Two rectilineal figures are said to be equiangular to
one another when the angles of the first, taken in order, are
equal respectively to those of the second, taken in order.

2. Eectilineal figures are said to be similar when they
are equiangular to one another, and also have the sides
about the equal angles taken in order proportionals.

Thus the two quadrilaterals A6CD, EFQH are similar if the
angles at A, 6, C, D are respec-
tively equal to those at E, F, G, H,
and if the following proportions

AB:BC:: EF : FG, ^*

BC : CD : : FG : GH,
CD.DA::GH: HE,
DAiAB:: ME: EF.

In these proportions, sides which are both antecedents or both
consequents of the ratios are said to be homologous or co7Tesponding.

[Def. 6, p. 320.]

Thus AB and EF are homologous sides ; so are BC and FG.

3. Two similar rectilineal figures are said to be similarly
situated with respect to two of their sides when these sides
are homologovs,

4. Two figures are said to have their sides about one
angle in each reciprocally proportional when a side of the
first figure is to a side of the second as the remaining side of
the second figure is to the remaining side of the first,

5. A straight line is said to be divided in extreme and
mean ratio when the whole is to the greater segment as
the greater segment is to the less.

326 EUCLID'S ELEMENTS.

Proposition 1. Theorem. [Euclid's Proof.]

The areas of triangles of the same altitude are to one another
as their bases.

Let ABC, ACD be two triangles of the same altitude,
namely the perpendicular from A to BD.

Then shall the A ABC : the A ACD :: BC : CD.

Produce BD both ways ;
and from CB produced cut oS way number of parts BG, GH,
each equal to BC ;

and from CD produced cut off any number of parts DK, KL,
LM, each equal to CD.

Join AH, AG, AK, AL, AM.

Since the A' ABC, ABG, AGH are of the same altitude, and
stand on the equal bases CB, BG, GH,

.*. the A' ABC, ABG, AGH are equal in area; L 38.
.*. the A AHC is the same multiple of the A ABC that HC
is of BC.

Similarly the A ACM is the same multiple of the A ACD
that CM is of CD.

And if HC = CM,

the A AHC = the A ACM ; L 38.

and if HC is greater than CM,
the A AHC is greater than the A ACM ; L 38, Cor.

and if HC is less than CM,
the A AHC is less than the A ACM. L 38, Cm-,

Now since there are four magnitudes, namely, the A" ABC,
ACD, and the bases BC, CD; and of the antecedents,
any equimultiples have been taken, namely, the A AHC

BOOfc Vt. PROP. 1. 327

and the base HC; and of the consequents, any equi-
multiples have been taken, namely the A ACM and the
base CM ; and since it has been shewn that the A AHC is
greater than, equal to, or less than the A ACM, according
as HC is greater than, equal to, or less than CM ;
.*. the four original magnitudes are proportionals ; v. Def, 5.
that is,
the A ABC : the A ACD :: the base BC : the base CD. •Q.E.D.

Corollary. The areas of parallelograms of the same
altitude are to one another as their bases.

Let EC, CF be par™" of the same altitude.

Then shall the par^ EC : the par^ CF :: BC : CD.

Join BA, AD.
Then the A ABC : the A ACD :: BC : CD ; Proved,
but the par"* EC is double of the A ABC, I. 34.
and the par™ CF is double of the A ACD ;
.-. the par™ EC : the par™ CF :: BC : CD. v. 8.

Note.

This proof of Proposition 1 is founded on Euclid's Test of Pro-
portion, and therefore holds good whether the bases BC, CD are
commensurable or otherwise.

The numerical treatment given on the following page applies in
strict theory only to the former case ; but the beginner would do
well to accept it, at any rate provisionally, and tnus postpone to
a later reading the acknowledged difficulty of Euclid's Theory of
Proportion.

327a

EUCLID'S ELEMENTS,

Proposition 1. [Numerical Illustration.]

The areas of triangles of equal aliUvde a/re to one another
as their bases,

A D

BLMNC ERSF

Let ABC, DEF be two triangles between the same par^
and therefore of equal altitude.

Then shall the A ABC : the A DEF =^ the base BC : the base EF.

Suppose BC contains 4 units of length, and EF 3 units ;
and let BL, LM, MN, NC each represent one unit, as also ER,
RS, SF.

Then BC:EF = 4:3.

Join AL, AM, AN ; also DR, DS.

Then the four A" ABL, ALM, AMN, ANC are all equal;
for they stand on equal bases, and are of equal altitude.

.-. the A ABC is four times the A ABL

Similarly, the A DEF is three times the A DER.

But the A' ABL and DER are equal, for they are on
equal bases BL, ER, and of equal altitude ;

hence the A ABC : the A DEF = 4 : 3

= BC : EF.

This reasoning holds good however many units of length
the bases BC, EF contain.

Thus if BO = m units, and EF = ?i units, then, whatever
whole numbers m and n represent,

the A ABC : the A DEF = m : n

= BC : EF.

The corollary should then be proved as on page 327.

BOOK VI. PROP. 1. 327b

Exercises on Proposition 1.

1. Two triangles of equal altitude stand on bases of 6*3 inches
and 5*4 inches respectively ; if the area of the first triangle is 12 j
square inches, find the area of the other. [10^ sq. in.]

2. The areas of two triangles of equal altitude have the ratio
24 : 17 ; if the base of the first is 4*2 centimetres, find the base of
the second to the nearest millimetre. [3*0 cm.]

3. Two triangles lying between the same parallels have bases of
16*20 metres and 20*70 metres ; find to the nearest square centimetre
the area of the second triangle, if that of the first is 50*1204 sq. metres.

[600427 sq. m.]

4. Assuming that tlie area of a triangle=\ hoM x altitude, prove
algebraically that

(i) Triangles of equal altitudes are proportional to their bases ;
(ii) Triangles on equal bases are proportional to their altitudes.

Also deduce the second of these propositions geometrically
from the first.

5. Two triangular fields lie on opposite sides of a common base ;
and their altitudes with respect to it are 4*20 chains and 3*71 chains.
If the first field contains 18 acres, find the acreage of the whole
quadrilateral. [33*9 acres.]

Definition.

Two straight lines are cut proportioiially when the segments of
one line are in the same ratio as the corresponding segments of the
other. [See definition, page 139.]

Fig. I. Fig. 2.

A X B A B X

Y D

Thus AB and CD are cut proportionally at X and Y, if

AX:XB::CY:YD.

And the same definition applies equally whether X and Y divide
AB and CD internally as in Fig. 1 or externally as in Fig. 2.

328

Euclid's elements.
Proposition 2. Theorem.

If a, straight line is drawn parallel to one side of a triangle^
it cuts the other sides, or those sides produced, proportionally.

Conversely, if the sides, or the sides produced, are ctU pro-
poi'tionally, the straight line which joins the points of section,
is parallel to the remaining side of the triangle.

Let XY be drawn par^ to BC, one of the sides of the
A ABC.

Then shall BX : XA :: CY : YA.

•

Join BY, ex.
Now the A" BXY, CXY are on the same base XY and
between the same par^' XY, BC ;

.-. the A BXY = the ACXY; L 37.

and AXY is another triangle ;
.-. the A BXY : the A AXY : : the A CXY : the A AXY. V. 4.
But the A BXY : the A AXY : : BX : X^ VI. 1.

and the A CXY : the A AXY :: CY : YA;

.-. BX:XA::CY:YA. V. 1.

Conversely. Let BX : XA : : CY : YA, and let XY be joined.
Then shall XY be par^ to BC.
As before, join BY, CX.
By hypothesis, BX : XA : : CY : YA ;

but BX : XA : : the A BXY : the A AXY, VI. I.
and CY : YA : : the A CXY : the A AXY ;
.-. the A BXY : the A AXY :: the A CXY : the A AXY. V. I,

,', the A BXY = the A CXY ; V. 6.

and these triangles are on the same base and on the same
side of it ;

.-. XY is par^ to BC. l, 39.

Q.E.D.

BOOK VI. PROP. 2. 329

EXERCISES.

1. Shew that every quadrilateral is divided by its diagonals
into four triangles whose areas are proportionals.

2. If any two straight lines are cut hy three parallel straight lines,
they are cut proportioTUjUly.

3. From the point E in the common base of two triangles ACB,,
AD6, straight lines are drawn parallel to AC, AD, meeting BC, BD
at F, G : shew that FQ is parallel to CD.

4. In a triangle ABC the straight line DEF meets the sides
BC, CA, AB at the points D, E, F respectively, and it makes equal
angles with AB and AC : prove that

BD :CD ::BF : CE.

5. In a triangle ABC, AD is drawn perpendicular to BD, the
bisector of the angle at B : shew that a straight line through D
parallel to BC will bisect AC.

6. From B and C, the extremities of the base of a triangle ABC,
straight lines BE, CF are drawn to the opposite sides so as to inter-
sect on the median from A : shew that EF is parallel to BC.

7. From P, a given point in the side AB of a triangle ABC,
draw a straight line to AC produced, so that it will be bisected
by BC.

8. Find a point within a triangle such that, if straight lines be
drawn from it to the three angular points, the triangle will be
divided into three equal triangles.

330 EUCLID'S ELEMENTS.

Proposition 3. Theorem.

If the vertical angle of a triangle he bisected by a straight
line which cuts the base, the segments of the base shall have to one
another the same ratio as the remaining sides of ike triangle.

Conversely y if the base be divided so that its segments have to
one another the same ratio as the remaining sides of the triangle,
the straight line drawn from the vertex to the point of section
shall bisect the vertical angle.

In the A ABC, let the l BAG be bisected by AX, which
meets the base at X.

Then shall BX : XC :: BA : AC.

Through C draw CE par* to XA, to meet BA produced
at E. L 31.

Then because XA and CE are par*,

.-. the L BAX = the int. opp. l AEC, I. 29.

and the l XAC = the alt. l ACE. I. 29.

But the L BAX = the l XAC ; Hyp.
.-. the z. AEC = the ^ ACE ;

.-. AC = AE. L 6.

Again, because XA is par* to CE, a side of the A BCE,

.-. BX : XC :: BA : AE ; VL 2.

that is, BX : XC : : BA : AC.

BOOK VI. PROP. 3. 331

Conversely. Let BX : XC : : BA : AC ; and let AX be joined.
Thm shall the l BAX = the l XAC.

For, with the same construction as before,

because XA is par^ to CE, a side of the A BCE,

.-. BX : XC :: BA : AE. VI. 2.

But, hy hypothesiSy BX : XC :: BA : AC ;

.-. BA : AE :: BA : AC; V. 1.

.-. AE = AC;
.'. the L ACE = the l AEC. I. 5.

But because XA is par^ to CE,
.-. the z. XAC = the alt. l ACE. I. 29.

and the ext. l BAX = the int. opp. l AEC ; i. 29.
.-. the ^ BAX = the z. XAC.

Q.E.D.

EXERCISES.

1. The side BC of a triangle ABC is bisected at D, and the
angles ADB, ADC are bisected by the straight lines DE, DF, meeting
AS, AC at E, F respectively : shew that EF is parallel to BC.

2. Apply Proposition 3 to trisect a given finite straight line.

3. If the line bisecting the vertical angle of a triangle is divided
into parts which are to one another as the base to the sum of the
sides, the point of division is the centre of the inscribed circle.

4. ABCD is a quadrilateral : shew that if the bisectors of the
angles A and C meet in the diagonal BD, the bisectors of the angles
B and D will meet on AC.

5. Construct a triangle liavtng given the bcue, the vertical angle,
and the ratio of the remaining sides,

6. Employ Proposition 3 to shew that the bisectors of the
angles of a triangle are concurrent.

7. AB is a diameter of a circle, CD is a chord at right angles to
it, and E any point in CD ; AE and BE are drawn and produced
to cut the circle in F and Q : shew that the quadrilateral CFDQ
has any two of its adjacent sides in the same ratio as the remaining
two.

332 EUCLID'S ELEMENTS.

Proposition A. Theorem.

If one side of a triangle be produced^ and the exterior angle
so formed he bisected by a straight line which cuts the base pro-
duced, the segments between the point of section and the extremities
of the base shall have to one another the same ratio as the
remaining sides of the triangle.

Conversely y if the segments of the base produced have to one
another the same ratio as the remaining sides of the triangle,
the straight line drawn from the vertex to the point of section
shall bisect the exterior vertical angle.

In the A ABC let BA be produced to F, and let the
exterior l CAF be bisected by AX which meets the base
produced at X.

Then shall BX : XC : : BA : AC.

Through C draw CE par' to XA, I. 31.

and let CE meet BA at E.

Then because AX and CE are par*,
/. the ext. L FAX = the int. opp. l AEC,

and the l XAC = the alt. l ACE. i. 29.

But the L FAX = the l XAC ; Hyp.

.'. the L AEC = the l ACE ;

.-. AC = AE. I. 6.

Again, because XA is par^ to CE, a side of the A BCE,

Constr.
.'. BX :XC :: BA: AE; VI. 2.

that is, BX : XC : : BA : AC.

BOOK VI. PROP. A. 333

Conversely, Let BX : XC : : BA : AC, and let AX be joined.
Then shall theLFfiX = theL XAC.

For, with the same construction as before,

because AX is par^ to CE, a side of the A BCE,

.-. BX : XC :: BA : AE. VI. 2.

But, by hypothesis, BX : XC : : BA : AC ;

.'. BA : AE :: BA : AC; V. 1.

.-. AE = AC;
.•. the L ACE = the l AEC. I. 5.

But because AX is par^ to CE,
.-. the L XAC = the alt. l ACE,
and the ext. l FAX = the int. opp. l AEC ; I. 29.
.*. the L FAX = the l XAC. Q.E.D.

Propositions 3 and A may be both included in one enunciation
as follows :

If the interior or exterior vertical angle of a triangle be bisected
by a straight line which also cvts the base, the base shall be divided
internally or externally into segments which have the same ratio as the
other sides of tlie triangle.

Conversely, if the base be divided internally or externally into seg-
ments which have the same ra^io as the other sides of the triangle, the
straight line drawn from the point of division to the vertex toill bisect
the interior or exterior vertical angle.

EXERCISES.

1. In the circumference of a circle of which AB is a diameter, a
point P is taken ; straight lines PC, PD are drawn eaually inclined
to AP and on opposite sides of it, meeting AB in C and D ;

shew that AC : CB :: AD : DB.

2. From a point A straight lines are drawn making the angles
BAC, CAD, DAE, each e^ual to half a right angle, and they are cut
by a straight line BCDE, which makes BAE an isosceles triangle :
shew that BC or DE is a mean proportional between BE and CD.

3. By means of Propositions 3 and A, prove that the straight
lines bisecting one angle of a triangle internally, and thQ other two
externally, are concurrent.

334 EUCLID'S ELEMENTS.

Proposition 4. Theorem.

If two triangles he equiangular to one anothery the sides about
the equal angles shall he proportionals^ those sides which are
opposite to e^ual angles heing homologous.

Let the A ABC be equiangular to the A DCE,
having the l ABC equal to the l DCE,
the L BCA equal to the l. CED,
and consequently the l CAB equal to the l EDC. I. 32.

Then shall the sides about these equal angles he proportionals^
namely

AB : BC :: DC : CE,

BC : CA :: CE : ED,

and AB : AC : : DC : DE.

Let the A DCE be placed so that its side CE may be
contiguous to BC, and in the same straight line with it.

Then because the z." ABC, ACB are together less than
two rt. angles, I. 17.

and the z. ACB = the z. DEC ; Hyp.

.*. the z."ABC, DEC are together less than two rt. angles;
.*. BA and ED will meet if produced. Ax. 12.
Let them be produced and meet at F.

Then because the l ABC = the z. DCE, Hyp,

.-. BF is par^ to CD ; L 28.

and because the z. ACB = the l DEC, Hyp,

.', AC is par^ to FE ; I. 28.

.-. FACD is a par"" ;

.*. AF = CD, and AC = FD. I. 34.

BOOK VI. PROP. 4. 335

Again, because CD is par^ to BF, a side of the A EBF,

.-. BC : CE :: FD : DE; VI. 2.

but FD = AC;
.-. BC : CE :: AC : DE ;

and, alternately, BC : CA :: CE : ED. V. 11

Again, because AC is par^ to FE, a side of the A FBE,

.-. BA : AF :: BC : CE ; VI. 2.

but AF = CD;,
.-. BA : CD :: BC : CE;

and, alternately, AB : BC .: DC : CE. V. 11.

Also BC : CA : : CE : ED j Proved,

.-., ex cequali, AB : AC :: DC : DE. V. 14.

Q.E.D.

[For Alternative Proof see Page 342.]

EXERCISES.

1. If one of the parallel sides of a trapezium is double the other,
shew that the diagonals intersect one another at a point of trisection.

2. In the side AC of a triangle ABC any point D is taken : shew
that if AD, DC, AB, BC are bisected in E, F, G, H respectively,
then EG is equal to HF.

3. AB and CD are two parallel straight lines ; E is the middle
point of CD ; AC and BE meet at F, and AE and BD meet at G :
shew that FG is parallel to AB.

4. ABCDE is a regular pentagon, and AD and BE intersect in
F : shew that AF : AE : : AE : AD.

5. In the figure of i. 43 shew that EH and GF are parallel, and
that FH and GE will meet on CA produced.

6. Chords AB and CD of a circle are produced towards B and
D respectively to meet in the point E, and through E, the line EF is
drawn parallel to AD to meet CB produced in F. Prove that EF is
a mean proportional between FB and FC.

xl.S.B.

336 EUCLID'S ELEMENTS

Proposition 5. Theorem.

If the sides of two triangles^ taken in m-der about earn of
their angles^ he proportionals^ the triangles shall be equiangular
to one another, having those angles equal which are opposite to
the homologous sides.

Let the A" ABC, DEF have their sides proportionals, so
that AB : BC :: DE : EF,

BC : CA :: EF ; FD,
and consequently, ex cequali,

AB : AC :: DE : DF.

Then shall the A" ABC, DEF be equiangular to one another.

At E in FE make the l FEG equal to the l ABC ; L 23.
and at F in EF make the ^ EFG equal to the l BCA ;

.*. the remaining l EGF = the remaining l BAC. I. 32.

Then the A" ABC, GEF are equiangular to one another ;

.-. AB : BC ::GE : EF. VL 4.

But, by hypothesis, AB : BC :: DE : EF ;

.'. GE : EF ::DE: EF; V. 1.

.-. GE = DE.
Similarly GF = DF.

Then in the A" GEF, DEF,

j GE = DE,

Because-! GF=DF,

[and EF is common j
.-. the L GEF = the l DEF, L 8.

and the l GFE = the l DFE,
and the l EGF = the l EDF.

BOOK VI. PROP. 5. 337

But the L GEF = the l ABC ; Ccmir,

.-. the L DEF = the L ABC.

Similarly, the l EFD = the l BCA;
.*. the remaining l FDE = the remaining l CAB; I. 32.
that is, the A DEF is equiangular to the A ABC.

Q.E.D.

NOTE ON SIMILAR FIGURES.

Similar figures may be described as those which have the %a/mt
shape.

For this, two conditions are necessary [see vi., Def, 2] ;

(i) the figures must have their angles equal each to each ;

(ii) their sides ahovi the equal angles taken in order must he
proportional.

In the case of triangles we have learned that these conditions
are not independent, for each follows from the other : thus

(i) if the triangles are equiangidar. Proposition 4 proves the
proportionality of their sides ;

(ii) if the triangles have their sides proportional. Proposition 5
proves their equiangvlarity.

This, however, is not necessarily the case
with rectilineal figures of more than three
sides. For example, the first diagram in the
margin shews two figures which are equi-
angular to one another, but which clearly
have not their sides proportional ; while the
figures in the second diagram have their sides
proportional, but are not equiangular to one
another.

338 EUCLID'S ELEMENTS.

Proposition 6. Theorem.

If two triangles have one atigle of the one equal to one angle
of the otheTy and the sides about the equal an^gles proportionals,
the triangles shall be similar.

In the A" BAG, EDF, let the l BAG = the l EDF,
and let BA : AG :: ED : DR.

Then shall the A" BAG, EDF be similar.

At D in FD make the l FDG equal to the z. CAB : I, 23.

at F in DF make the l DFG equal to the l ACB ;
.-. the remaining z. DGF = the remaining z. ABG. L 32.

Then the A" BAG, GDF are equiangular to one another ;

.*. BA : AG :: GD : DF. VI. 4.

But, by hypothesis, BA : AG :: ED : DF ;

.*. GD : DF :: ED : DF,
.-. GD=ED.

Then in the A' GDF, EDF,
I GD = ED,

Because i' and DF is common ;

[and the l GDF = the l EDF ; Constr,

.*. the A* GDF, EDF are equal in all respects ; I. 4.
so that the A EDF is equiangular to the A GDF ;
but the A GDF is equiangular to the A BAG ; Constr.
.'. the A EDF is equiangular to the A BAG;
/. their sides about the equal angles are proportionals; VI. 4.
that is, the A" BAG, EDF are similar.

Q.E.D.

BOOK VI. PROP. 6. 339

EXERCISES.
ON Propositions 1 to 6.

1. Shew that the diagonals of a trapezium cut one another in
the same ratio.

2. If three straight lines drawn from a point cut two parallel
straight lines in A, B, C and P, Q, R respectively, prove that

AB : BC :: PQ : QR.

3. From a point O, a tangent OP is drawn to a given circle, and
a secant OQR is drawn cutting it in Q and R ; shew that

OQ : OP :: OP : OR.

4. If two triangles are on equal bases and between the same
parallels, any straight line parallel to their bases vnll cut off equal
areas from the two triangles,

5. If two straight lines PQ, XY intersect in a point O, so that
PO : OX : : YO : OQ, prove that P, X, Q, Y are concyclic,

6. On the same base and on the same side of it two equal
triangles AOB, ADB are described ; AC and BD intersect in O, and
through O lines parallel to DA and CB are drawn meeting the base
in E and F. Shew that AE = BF.

7. BD, CD are perpendicular to the sides AB, AC of a triangle
ABC, and CE is drawn perpendicular to AD, meeting AB in E : shew
that the triangles ABC, ACE are similar.

8. AC and BD are drawn perpendicular to a given straight line
CD from two ^ven points A and d ; AD and BC intersect in E, and
EF is perpendicular to CD : shew that AF and BF make equal angles
with CJD.

9. ABCD is a parallelogram ; P and Q are points in a straight
line parallel to AB ; PA and QB meet at R, and PD and QC meet at
S : shew that RS is parallel to AD.

10. In the sides AB, AC of a triangle ABC two points D, E are
taken such that BD is equal to CE ; if DE, BC produced meet at F,
shew that AB : AC : : EF : DF.

11. Find a point the perpendiculars from which on the sides of
a given triangle shall be in a given ratio.

340 Euclid's elements.

Proposition 7. Theorem.

If two triangles have one angle of the one equal to one angle
of the other, and the sides about one other angle in each propor-
tional, so that the sides opposite to the equal angles a/re homologous,
then the third angles are either equal or supplementary ; and in
the former case the triangles are similar.

Let ABC, DEF be two triangles having the l ABC equal
to the z. DEF, and the sides about the angles at A and D
proportional, namely

BA : AC :: ED : DF.
Then shall the l' ACB, DFE be either equal (as in Figs, 1 and 2)
or supplementary (as in Figs, 1 and 3), and in the farmer
case the triangles shall be similar.

If the L BAG = the l EDF, [Figs, 1 and 2.]

then the l ACB = the l DFE ; I. 32.

and the A' are equiangular, and therefore similar, vi. 4.

But if the L BAC is not equal to the l EDF, [Figs, 1 and 3.]

one of them must be the greater.
Let the z_ EDF be greater than the l BAC.
At D in ED make the l EDF' equal to the l BAC. [Fig. 3.]
Then the A' BAC, EDF' are equiangular, I. 32.

. * . B A ! AC

but, by hypothesis, BA : AC

.-. ED : DF

: ED : DF'; VL 4.

ED : DF;

ED : DF', V. 1.

.-. DF = DF',
.-. the L DFF' = the l DF'F. L 5.

But the L" DF'F, DF'E are supplementary, I. 13.
.*. the z_' DFF', DF'E are supplementary :
that is, the z_' DFE, ACB are supplementary. Q.E.D.

BOOK VI. PROP. 7. 341

Corollaries to Proposition 7.

Three cases of this theorem deserve special attention.

It has been proved that if the angles ACB, DFE are not supple-
mentary ^ they are equaZ,

Hence, in addition to the hypothesis of this theorem,

(i) If the angles ACB, DFE, opposite to the two homo-
logous sides AB, DE are both acute or both obtuse, they
cannot be supplementary, and are therefore equal : or
if one of them is a right angle, the other must also be
a right angle (whether considered as supplementary or
equal to it) :

in either case the triangles are similar.

(ii) If the two given angles at B and E are right angles or
obtuse angles, it follows that the angles ACB, DFE
must be both acute, and therefore equal, by (i) :
so that the triangles are similar.

(iii) If in each triangle the side opposite the given ancle is
not less than the other given side ; that is, if AC and
DF are not less than AB and DE respectively, then
the angles ACB, DFE cannot be greater than the angles
ABC, DEF, respectively;

therefore the angles ACB, DFE, are both acute ;
hence, as above, they are equal ;
and the triangles ABC, DEF are similar.

342 EUCLID'S ELEMENTS.

Ohs, We have given Euclid's demonstrations of Propositions 4,
5, 6 ; but these propositions also admit of easy proof by the method
of superposition.

As an illustration, we will apply this method to Proposition 4.

Proposition 4. [Alternative Proof.]

If tioo triangles he equiangvlar to one, another, the sides about the
equal angles shidl he proportionals, those sides which are opposite to
equal angles heing homologous.

Let the A ABC be equiangular to the A DEF,
having the L ABU equal to the Z. DEF,
the L BCA equal to the L EFD,
and consequently the L CAB equal to the Z. FDE. i. 32.

Then shall the sides about these equcd angles he proportionals.

Apply the A ABC to the A DEF, so that B falls on E, and BA.
along ED :

then BC will fall along EF, since the L ABC = the L DEF. Hyp,
Let Q and H be the points in ED and EF, on which A and C faU;
then GH represents AC in its new position.

Then because the L EGH {i.e. the L BAC) = the L EDF, Hyp.

.'. GH is pari to DF : vi. 2.

.-. DG:GE::FH : HE;
.*. , componendo, DE : GE :: FE : HE, v. 13.

.*., alternately, DE : FE :: GE : HE, v. 11.

that is, DE : EF :: AB : BC.

Similarly by applying the A ABC to the A DEF, so that the
point C may fall on F, it may be proved that

EF : FD ::BC : CA.
.*., ex (xqudli, DE : DF : : AB : AC.

Q.E.D.

QUESTIONS FOR REVISION. 343

QUESTIONS FOR REVISION, AND NUMERICAL ILLUSTRATIONS.

1. Distinguish between the use of the word equiatigular in the
following cases :

(i) the figure ABCD is equiangular ;

(ii) the figure ABCD is equiangular to the figure EFGH.

2. Define the terms ratio, antecedent, consequent. Why must
the terms of a ratio be of the same kiiid ? When are ratios said to
be reciprocal ?

3. When are four quantities in proportion 1 Quote the alge-
braical and geometrical tests of proportion ; and deduce the latter
from the former.

4. What is meant by homologous terms in a proportion ? In the
enunciation of Prop. 4, why is it necessary to add — those sides which
are opposite to equal angles being homologous ?

5. Quote the enunciation of the theorem known as altemando
or alternately ; and explain why the terms of a proportion to which
this theorem is applied must be all of the same kind.

6. In the Particular Enunciation of Proposition 5 it is given
that AB : BC :: DE : EF,

and BC : CA :: EF : FD;

Why do we add **a?id consequently"

AB :CA ::DE:FD?

7. Define similar figures. In what way do the conditions of simi-
larity in triangles differ from those in figures of more than three sides?

8. Two parallelograms whose areas are in the ratio 2*1 : 3*5 lie
between the same parallels. If the base of the first is 6*6 inches in
length, shew that the base of the second is 11 inches.

9. ABC is a triangle, and XY is drawn parallel to BC, cutting
the other sides at X and Y :

(i) If AB = 1 foot, AC = 8 inches, and AX = 7 inches; shew that

AY=4§ inches,
(ii) If AB=20 inches, AC = 15 inches, and AY = 9 inches, shew

that BX = 8 inches,
(iii) If X divides AB in the ratio 8 : 3, and if AC =2*2 inches,
shew that AY, YC measure respectively 1 '6 and '6 inches.

10. The vertical angle A of a triangle ABC is bisected by a line
which cuts BC at X ; if BC = 25 inches in length, and if the sides
BA, AC are in the ratio 7 : 3, shew that the segments of the base
are 17*5 and 7*5 inches respectively.

344 euclid's elements.

Proposition 8. Theorem.

In a right-angled triangle, if a perpendicular is drawn from
the right angle to the hypotenuse, the triangles on each side of it
are similar to the whole triangle and to one another.

B DC

Let BAC be a triangle right-angled at A, and let AD be
drawn perp. to BC.

Then shall the A' BDA, ADC be similar to the A BAC and
to one another.

In the A» BDA, BAC,

the L BDA = the i. BAC, being rt. angles,

and the angle at B is common to both ;

.'. the remaining l. BAD = the remaining l BCA, I. 32.

that is, the A BDA is equiangular to the A BAC ;

.-. their sides about the equal angles are proportionals; VI. 4.

.-. the A" BDA, BAC are similar.
In the same way it may be proved that the A" ADC,
BAC are similar.

Hence the A' BDA, ADC, having their angles severally
equal to those of the A BAC, are equiangular to one another;

.'. they are similar. VL 4.

Q.E.D.

Corollary. Because the A» BDA, ADC are similar,

.-. BD : DA : : DA : DC ;

and because the A" CBA, ABD are similar,
.-. CB : BA : : BA : BD ;

and because the A' BCA, ACD are similar,
.-. BC : CA : : CA : CD.

EXERCISES.

1. In the figure of Prop. 8 prove that the hypotenuse is to one
side as the second side is to the perpendicular.

2. Shew that the radius of a circle is a mean proportional between
the segments of any tangent between its point of contact and a pair
of parallel tangents.

BOOK VI. PROP. 9. 345

Definition. One magnitude is said to be a submultiple
of another, when the first is contained an eoixtct number of
times in the second. [Book v. Def. 2.]

Proposition 9. Problem.
From a given straight line to cut off any required svhmuUiple.

A F

Let AB be the given straight line.
It is required to cut off a certain sfuhmvltiple from AB.

From A draw a straight line AG of indefinite length, making
any angle with AB.

In AG take any point D ; and, by cutting off successive
parts each equal to AD, make AE to contain AD as many
times as AB contains the required submultiple.

Join EB.
Through D draw DF par^ to EB, meeting AB in F.

Then shall AF he the required submultiple.

Because DF is par^ to EB, a side of the A AEB,

.-. BF : FA :: ED : DA; VI. 2.

.'., componendo, BA : AF :: EA : AD. V. 13.

But AE contains AD the required number of times ; Constr.
.*. AB contains AF the required number of times ;
that is, AF is the required submultiple. Q.E.F.

EXERCISES.

1. Divide a straight line into five equal parts.

2. Give a geometrical construction for cutting off two-sevenths
of a given straight line.

346 EUCLID^S ELEMENTS.

Proposition 10. Problem.

To divide a straight line similarly to a given divided straight
line.

B K

Let AB be the given straight line to be divided, and AC
the given straight line divided at the points D and E.

It is required to divide AB similarly to AC.

Let AB, AC be placed so as to form any angle.

Join CB.
Through D draw DF par* to CB, T. 31.

and through E draw EG pai^ to CB.

Then AB shall be divided at F and G similarly to AC.

Through D draw DHK par* to AB.
Now by construction each of the figs. FH, HB is a par* ;

.-. DH = FG, and HK = GB. L 34.

Now since HE is par^ to KC, a side of the A DKC,

.-. KH : HD :: CE : ED. VL 2.

But KH = BG, and HD = GF;

.-. BG : GF :: CE : ED. V. 1.

Again, because FD is par^ to GE, a side of the A AGE,

.-. GF : FA :: ED : DA; VL 2.

.*. , ex cequaliy BG : FA :: CE : DA: V. 14.

.'. AB is divided similarly to AC, Q.E.F.

exercise.

Divide a straight line internally and externally in a given raiio.
Is this alioays possible ?

book vi. prop. 11. 347

Proposition 11. Problem.
To find a thM proportional to two given straight lines.

Let A, B be two given straight lines.
It is required to find a thii'd proportional to A and B.

Take two st. lines DL, DK of indefinite length, containing
any angle.

From DL cut off DG equal to A, and GE equal to B ;

and from DK cut off DH also equal to B. I. 3.

Join GH.
Through E draw EF par^ to GH, meeting DK in F. I. 31.

Then shall HF be a third proportional to A and B.

Because GH is par^ to EF, a side of the A DEF ;

.-. DG : GE :: DH : HF. VI. 2.

But DG = A; and GE, DH each = B; Constr,

.*• A I D II D I HP j

that is, H F is a third proportional to A and B.

Q.E.F.

EXERCISES.

1. AB is a diameter of a circle, and through A any straight line
is drawn to cut the circumference in C and the tangent at d in D :
shew that AC is a third proportional to AD and AB.

2. ABC is an isosceles triangle having each of the angles at the
base double of the vertical angle BAC ; the bisector of the angle BCA
meets AB at D. Shew that AB, BC, BD are three proportionals.

3. Two circles intersect at A and B ; and at A tangents are
drawn, one to each circle, to meet the circumferences at C and D :
shew that if CB, BD are joined, BD is a third proportional to CB,
BA.

348 Euclid's elements.

Proposition 12. Problem.
To find a fourth proportional to three given straight lines.

ABC

Let A, B, C be the three given straight linea
It is required to find a fourth proportional to A, B, C.

Take two straight lines DL, DK of indefinite length, con-
taining any angle.
From DL cut off DG equal to A, and GE equal to B ;

and from DK cut off DH equal to C. I. 3.

Join GH.
Through E draw EF par^ to GH. I. 31.

Then shall HF be a fourth proportional to A, B, C.

Because GH is par* to EF, a side of the A DEF;

.-. DG : GE :: DH : HF. VL 2.

But DG==A, GE = B, and DH = C; Canstr.

.-. A : B :: C : HF;
that is, HF is a fourth proportional to A, B, C.

Q.E.F.

exercises.

1. If from D, one of the angular points of a parallelogram
ABCD, a straight line is drawn meeting AB at E and CB at F ;
shew that CF is a fourth proportional to EA, AD, and AB.

2. In a triangle ABC the bisector of the vertical angle BAG
meets the base at D and the circumference of the circumscribed
circle at E : shew that BA, AD, EA, AC are four proportionals.

3. From a point P tangents PQ, PR are drawn to a circle whose
centre is C, and QT is drawn perpendicular to RC produced : shew
that QT is a fourth proportional to PR, RC, and RT.

book vi. prop. 13. 349

Proposition 13. Problem.
To find a mean proporimial between two given straight lines.

A B C

Let AB, BC be the two given straight lines.
It is required to find a mean proportional between AB and BC.

Place AB, BC in a straight line, and on AC describe the
semicircle ADC.

From B draw BD at rt. angles to AC. I. 11.

Then shall BD be a mean propoiiional between AB and BC.

Join AD, DC.

Now the l ADC, being in a semicircle, is a rt. angle; ill. 31.
and because in the right-angled A ADC, DB is drawn from
the rt. angle perp. to the hypotenuse,

.'. the A' ABD, DBC are similar; VI. 8.

. . AB : BD :: BD : BC;
that is, BD is a mean proportional between AB and BC.

Q.E.F.

EXERCISES.

1. If from one angle A of a parallelogram a straight line is
drawn cutting the diagonal in E and the sides in P, Q, shew that
AE is a mean proportional between PE and EQ.

2. A, B, C are three points in order in a straight line : find a
point P in the straight line so that PB may be a mean proportional
between PA and PC.

3. The diameter AB of a semicircle is divided at any point C,
and CD is drawn at right angles to AB meeting the circumference
in D ; DO is drawn to O the centre, and CE is perpendicular to OD :
shew that DE is a third proportional to AG and DC.

350 Euclid's elements.

4. AC is the diameter of a semicircle on which a point B is
taken so that BC is equal to the radius : shew that AB is a mean
proportional between BC and the sum of BC, CA.

5. A is any point in a semicircle on BC as diameter ; from D
any point in BC a perpendicular is drawn meeting AB, AC, and the
circumference in E, U, F respectively ; shew that DG is a third
proportional to DE and DF.

6. Two circles have external contact, and a common tangent
touches them at A and B : prove that AB is a mean proportional
between the diameters of the circles. [See Ex. 21, p. 237.]

7. If a straight line is divided at two given points, determine
a third point such that its distances from the extremities may be
proportional to its distances from the given points.

8. AB is a straight line divided at C and D so that AB, AC, AO
are in continued proportion ; from A a line AE is drawn in any
direction and equal to AC ; shew that BC and CD subtend equal
angles at E.

9. In a given triangle draw a straight line parallel to one of the
sides, so that it may be a mean proportional between the segments
of the base.

10. On the radius OA of a quadrant OAB, a semicircle ODA is
described, and at A a tangent At is drawn ; from O any line ODFE
is drawn meeting the circumferences in D and F and the tangent in
E : if DG is drawn perpendicular to OA, shew that OE, Or , OD,
and OG are in continued proportion.

11. From any point A, on the circumference of the circle ABE,
as centre, and with any radius, a circle BDC is described cutting
the former circle in B and C ; from A any line AF E is drawn meeting
the chord BC in F, and the circumferences BDC, ABE in D, E
respectively : shew that AD is a mean proportional between AF
and AE.

Definition. Two figures are said to have their sides
about one angle in each reciprocally proportional, when a
side of the first figure is to a side of the second as the
remaining side of the second figure is to the remaining side
of the first. [Book vi. Def. 4.]

book vi. prop. 14. 351

Proposition 14. Theorem.

Parallelograms which are equal in area, and which have one
angle of the one equal to one angle of the other, have their sides
about the equal angles reciprocally proportional.

Conversely, parallelograms which have one angle of the one
equal to one angle of the other, and the sides about these angles
recipvcally propmiional, are equal in area.

A F

G C

Let the par"" AB, BC be of equal area, and have the
A. DBF equal to the L QBE.

Then shall the sides abmit the l' DBF, GBE he reciprocally
proportional,
namely, DB : BE :: GB : BF.

Place the par™" so that DB, BE may be in the same straight
line;

.*. FB, BG are also in one straight line. I. 14.
Complete the par™ FE.

Then because the par"" AB = the par™ BC, Hyp.
and FE is another par™,
.-. the par™ AB : the par™ FE :: the par™ BC : the par™ FE ;
but the par™ AB : the par™ FE :: DB : BE, VI. 1. Cm-,

and the par™ BC ; the par™ FE : : GB : BF ;

.*. DB : BE :: GB : BF. V. 1.

Conversely. Let the z_ DBF be equal to the l GBE,

and let DB : BE : : GB : BF.
Then shall the par^ AB he equal in area to the pair^ BC.

For, with the same construction as before,
by hypothesis, DB : BE :: GB : BF;

but DB : BE : : the par™ AB : the par™ FE, VI. 1.
and GB : BF : : the par™ BC : the par™ FE,
.'.the par™ AB : the par™ FE : : the par™ BC : the par™ FE; V.L

.-. the par™ AB = the par™ BC. Q.E.D.

H.S.E. Z

352 euclid's elements.

Proposition 15. Theorem.

Triangles which are equal in area, and which have ane angle
of the one equal to one angle of the other, have their sides about
the equal angles reciprocally pi'opmiianal.

Conversely, triangles which have one angle of the one equal
to one angle of the other, and the sides about these angles re-
cip'ocally propoiiional, are equal in area,

Let the A" CAB, EAD be of equal area, and have the
L CAB equal to the l EAD.

Then shall the sides about the l* CAB, EAD be reciprocally
proportional,
namely, CA : AD :: EA : AB.

Place the A" so that CA and AD may be in the same st. line;
.*. BA, AE are also in one st. line. i. 14.

Join BD.

Then because the A CAB = the A EAD, Hyp.

and ABD is another triangle,
.-. the A CAB : the A ABD
but the A CAB : the A ABD

and the A EAD ; the A ABD

.-. CA ! AD

: the A EAD : the A ABD ;
: CA : AD, VL 1.

: EA : AB ;
: EA : AB. V. 1.

Conversely. Let the l CAB be equal to the l EAD,
and let CA : AD : : EA : AB.

Then shall the A CAB = the A EAD.

For, with the same construction as before,
by hypothesis, CA : AD : : EA : AB ;

but CA : AD :: the A CAB : the A ABD, VL 1.
and EA : AB : : the A EAD : the A ABD ;
.-. the A CAB : the A ABD :: the A EAD : the A ABD ; V. 1,
.*. the A CAB = the A EAD. Q.E.D.

BOOK VI. PROP. 15. 353

EXERCISES.

ON Propositions 14 and 15.

1. Parcdlelograms which are equal in area and which have their
sides reciprocally proportional, have their angles respectively equal,

2. Triangles which are equal in area, and which have the sides
about a pair of angles reciprocally proportional^ have those angles
equal or supplementary,

3. AC, BD are the diagonals of a trapezium which intersect in
O ; if the side AB is parallel to CD, use Prop. 15 to prove that the
triangle ADD is equal to the triangle BOC.

4. From the extremities A, B of the hypotenuse of a right-
angled triangle ABC lines AE, BD are drawn perpendicular to AB,
and meeting BC and AC produced in E and D respectively : employ
Prop. 15 to shew that the triangles ABC, ECD are equal in area.

5. On AB, AC, two sides of any triangle, squares are described
externally to the triangle. If the squares are ABDE, ACFG, shew
that the triangles DAG, FAE are equal in area.

6. ABCD is a parallelogram ; from A and C any two parallel
straight lines are drawn meeting DC and AB in E and F respectively ;
EG, which is parallel to the diagonal AC, meets AD in G: shew that
the triangles DAF, GAB are equal in area.

7- Describe an isosceles triangle equal in area to a given triangle
and having its vertical angle equal to one of the angles of the given
triangle.

8. Prove that the equilateral triangle described on the hypo-
tenuse of a right-angled triangle is equal to the sum of the equilateral
triangles described on the sides containing the right angle.

[Let ABC be the triangle right-angled at C ; and let BXC, CYA,
AZB be the equilateral triangles. Draw CD perpendicular to AB ;
and join DZ. Then shew by Prop. 15 that the A AYC = the A DAZ ;
and similarly that the A BXC = the A BDZ.]

354

EUCLID'S ELEMENTS.

Proposition 16. Theorem.

If four straight lines are proportional, tlie rectangle con-
tained by the extremes is equal to the rectangle contained by the
means.

Conversely, if the rectangle contained by the extremes is equal
to the rectangle contained by the means, the four straight lines
are propmiional.

E G

Let the st. lines AB, CD, EF, GH be proportional, so that

AB : CD :: EF : GH.

Then shall the red AB, GH = the red. CD, EF.

From A draw AK perp. to AB, and equal to GH. L 11, 3.
From C draw CL perp. to CD, and equal to EF.
Complete the par™' KB, LD.

Then because AB : CD : : EF : GH ; Hijp.

and EF = CL, and GH = AK ; Constr.

.-. AB : CD :: CL : AK;
that is, the par™' KB, LD have their sides about the equal
angles at A and C reciprocally proportional ;

.-. KB=LD. VL 14.

But KB is the rect. AB, GH, for AK = GH, Constr.
and LD is the rect. CD, EF, for CL= EF;

.-. the rect. AB, GH = the rect. CD, EF.

BOOK VI. PROP. 16. 365

Conversely, Let the rect. AB, GH = the rect. CD, EF.

Then shall AB : CD :: EF : GH.
For, with the same construction as before,

because the rect. AB, GH = the rect. CD, EF ; Hyp,
and the rect. AB, GH = KB, for GH = AK, Caiistr,
and the rect. CD, EF= LD, for EF = CL;

.'. KB=LD;
that is, the par"" KB, LD, which have the angle at A equal
to the angle at C, are equal in area ;
.*. the sides about the equal angles are reciprocally
proportional ;

that is, AB : CD :: CL : AK ;
.-. AB : CD :: EF : GH.

Q.E.D.

QUESTIONS FOR REVISION.

1. State and prove the algebraical theorem corresponding to
Proposition 16.

2. Define the terms : multiple, suhmultiplef fourth proportionalj
third proportional, mean proportional.

3. ABC is a triangle right-angled at A, and AD is drawn per-
pendicular to BC : if AB, AC measure respectively 12 and 5 inches,

shew that the segments of the hypotenuse are lliV ^/nd \\% inches.

4. Find in inches the length of the mean proportional between
1 inch and 3 inches. Hence give a geometrical construction for

drawing a line >/3 inches in length : and extend the method to

finding a line fjn inches long.

5. A straight line AB, 21 inches in length, is divided at F and
G into parts of 5, 7, 9 inches respectively. If a second line AC, 35
inches long, is similarly divided by the method of Proposition 10,

shew that the lengths of the parts are 8^, 11§ and 15 inches re-
spectively.

6. When are figures said to have their sides about one angle in
each reciprocally proporti&iml ? Two equal parallelograms ABCD,
EFGH have their angles at B and F equal : if AB=2 inches, BC=10
inches, and EF=5 inches; find the length of FG.

356

Euclid's elements.

Proposition 17. Theorem.

If three straight lines are p'opmiional the rectangle contained
by the extremes is equal to the square on the mean.

Conversely, if the rectangle contained by the extremes is equal
to the square on the mean, the three straight lines are propor-
tional.

•C
■D
■B
•A

VL 16.

Let the three st. lines A, B, C be proportional, so that

A : B : : B : C.
Then slvall the reel. A, C be equal to the sq. on B.

Take D equal to B.

Then because A : B : : B : C, and D = B ;
.-. A : B :: D : C;
.-. the rect. A, C = the rect. B, D;
but the rect. B, D = the sq. on B, for D = B ;
.'. the rect. A, C = the sq. on B.

Conversely. Let the rect. A, C = the sq. on B.

Then shall A : B : : B : C.
For, with the same construction as before,
because the rect. A, C = the sq. on B,

and the sq. on B = the rect. B, D, for D = B ;
.*. the rect. A, C = the rect. B, D;

.-. A : B :: D : C, VL 16.

that is, A : B : : B : C. Q.E.D.

Hfp.

QUESTIONS FOR REVISION.

1. State and prove the algebraical theorem corresponding to
Proposition 17.

2. Two adjacent sides of a rectangle measure 12*1 and '9 inches
in length ; shew that the side of an equal square is 3*3 inches.

3. ABC is an isosceles triangle, the equal sides each measuring
12 inches. DAE is a triangle of equal area, having the angle DAE
equal to the angle CAB. If AD = 36 inches, find the length of AE.

EXERCISES ON PROPS. 16 AND 17. 357

EXERCISES.
ON Propositions 16 and 17.

1. Apply Proposition 16 to prove that if two chords of a circle
intersect, tne rectangle contained by the segments of the one is equal
to the rectangle contained by the segments of the other.

2. Prove that the rectangle contained by the sides of a right-
angled triangle is equal to the rectangle contained by the hypotenuse
and the perpendicular drawn to it from the right angle.

3. On a given straight line construct a rectangle equal to a given
rectangle.

4. ABCD is a parallelogram ; from B any straight line is drawn
cutting the diagonal AC at F, the side DC at G, and the side AD
produced at E : shew that the rectangle EF, FG is equal to the
square on BF.

5. On a given straight line as base describe an isosceles triangle
equal to a given triangle.

6* AB is a diameter of a circle, and any line ACD cuts the circle
in C and the tangent at B in D ; shew by Prop. 17 that the rectangle
AC, AD is constant.

7. The exterior angle at A of a triangle ABC is bisected by a
straight line which meets the base in D and the circumscribed circle
in E : shew that the rectangle BA, AC is equal to the rectangle EA, AD.

8. If two chords AB, AC drawn from any point A in the cir-
cumference of the circle ABC are produced to meet the tangent at the
other extremity of the diameter through A in D and E, shew that
the triangle AED is similar to the triangle ABC.

9. At the extremities of a diameter of a circle tangents are
drawn ; these meet the tangent at a point P in Q and R : shew that
the rectangle QP, PR is constant for all positions of P.

10. A is the vertex of an isosceles triangle ABC inscribed in a
circle, and ADE is a straight line which cuts the base in D and the
circle in E ; shew that the rectangle EA, AD is equal to the square
on AB.

11. Two circles touch one another externally at A ; a straight
line touches the circles at B and C, and is produced to meet the
straight line joining the centres at S : shew that the rectangle SB, SC
is equal to the square on SA.

12. Divide a triangle into two equal parts by a straight line
drawn at right angles to one of the sides.

358 EUCLID^S ELEMENTS.

Definition. Two similar rectilineal figures are said to
be similarly situated with respect to two of their sides
when these sides are homologous, [Book vi. Def. 3.]

Proposition 18. Problem.

On a given straight line to describe a rectilineal figure similar
and similarly siituited to a given rectilineal figure.

A BCD

Let AB be the given st. line, and CDEF the given recti-
lineal figure.

It is required to descnbe on the st. line AB a rectilineal
figure similar and similarly situated to CDEF.

First suppose CDEF to be a quadrilateral.

Join DF.
At A in BA make the l BAG equal to the l DCF, I. 23.
and at B in AB make the l ABG equal to the l CDF ;

.*. the remaining l AGB = the remaining l CFD ; L 32.
and the A AG B is equiangular to the A CFD.

Again at B in GB make the l GBH equal to the l FDE,
and at G in BG make the ^ BGH equal to the l DFE ; I. 23.
.*. the remaining l BHG = the remaining z_ DEF; I. 32.
and the A BHG is equiangular to the A DEF.

Then shall ABHG he the required figure.

(i) To prove that the fig. ABHG is equiangular to the
fig. CDEF.

Because the l AGB = the l CFD, Proved.

and the l BGH = the l DFE ; Constr.

.'. the whole l AGH = the whole l CFE.
Similarly the l ABH = the L CDE ;
and the angles at A and H are respectively equal to the
angles at C and E ; Constr. and poof.

.*. the fig. ABHG is equiangular to the fig. CDEF.

BOOK VI. PROP. 18. 359

(ii) To prove that the figs. ABHG, CDEF have the sides
about their equal angles proportional.

Because the A AG B is equiangular to the A CFD,

.*. AG : GB :: CF : FD. VI. 4.

And because the A BGH is equiangular to the A DFE,

.-. BG : GH :: DF : FE;
.'., ex ceguali, AG : GH :: CF : FE. V. 14.

Similarly it may be shewn that

CD : DE.

DC : CF, VI. 4.

FE : ED.

AB : BH :
Also BA : AG :
and GH : HB :

.*. the figs. ABHG, CDEF are equiangular and have their
sides about the equal angles proportional ;

that is, ABHG is similar to CDEF. vi. Def 2.

In like manner the process of construction may be
extended to a figure of five or more sides.

Q.E.F.

Definition. When three magnitudes are proportionals
the first is said to have to the third the duplicate ratio of
that which it has to the second, [Book v. Def 13.]

360 Euclid's elements.

Proposition 19. Theorem.

Similar triangles are to one another in the dicplicate ratio of
their homologous sides,

Let ABC, DEF be similar triangles, having the L ABC
equal to the £. DEF, and let BC and EF be homologous sides.

Then shall the A ABC be to the A DEF in the duplicate ratio of
BC to EF.

To BC and EF take a third proportional BG,

so that BC : EF :: EF : BG. VL 11.

Join AG.

Then because the A" ABC, DEF are similar, Hyp,
.-. AB : BC :: DE : EF ;
.-., alternately, AB : DE :: BC : EF; V. 11.

but BC : EF :: EF : BG ; Cmistr.

,'. AB ; DE :: EF : BG ; V. 1.

that is, the sides of the A" ABG, DEF about the equal
angles at B and E are reciprocally proportio^ial ;

.-. the A ABG = the A DEF. VL 15.

Again, because BC : EF :: EF : BG, Consir,

.'. BC : BG in the duplicate ratio of BC to EF. v. Def, 13.
But the A ABC : the A ABG :: BC : BG ; VI. 1.

.*. the A ABC : the A ABG in the duplicate ratio

of BC to EF: V. 1.

and the A ABG = the A DEF; Proved,

.'. the A ABC : the A DEF in the duplicate ratio

of BC : EF. Q.E.D.

QUESTIONS FOR REVISION. 361

QUESTIONS FOR REVISION, AND NUMERICAL ILLUSTRATIONS.

1. Quote the Geometrical and Algebraical definitions of the
duplicate of the ratio a : b ; and deduce the latter from the former.
Estimate numerically the duplicate of the ratio 36 : 21.

2. The smaller of two similar triangles has an area of 20 square
feet, and two corresponding sides are 3 ft. 6 in. and 2 ft. 4 in. respec-
tively : shew that the area of the greater triangle is 45 square feet.

3. XY is drawn parallel to BC, the base of a triangle ABC, to
meet the other sides at X and Y : if AX and XB measure respec-
tively 3 inches and 7 inches, shew that the areas of the triangles
AXY, ABC are in the ratio 9 : 100.

4. Two similar triangles have areas in the ratio 529 : 361 ; shew
that any pair of homologous sides are to one another as 23 : 19.

5. When are similar figures said to be similarly aituatedt Shew
that similar and similarly situated triangles are to one another in the
duplicate ratio of their altitudes.

6. Two similar and similarly situated triangles have areas in
the ratio 1369 : 1681 ; if the altitude of the greater is 10 ft. 3 in.,
shew that the altitude of the other is 1 foot less.

7. The sides of a triangle are 11, 23, 29 ; find the sides of a
similar triangle whose area is 289 times that of the former.

8. Shew how to draw a straight line XY parallel to BC the base
of a triangle ABC, so that the area of the triangle AXY may be nine-
sixteenths of that of the triangle ABC.

9. XY is drawn parallel to the base BC of a triangle ABC, so
that the triangle AXY has to the figure XBCY the ratio 4:5; shew
that AB and AC are cut by XY in the ratio 2:1.

10. A triangle ABO is bisected by a straight line XY drawn
parallel to the base BC. In what ratio is AB divided at X ?

Hence shew how to bisect a triangle by a straight line drawn
parallel to the base.

11. ABC is a triangle whose area is 16 square feet; and XY
is drawn parallel to BC, dividing AB in the ratio 3:5; shew that if
BY is joined, the area of the triangle BXY is 3 sq. ft. 108 sq. in.

12. ABC is a triangle right-angled at A, and AD is the perpen-
dicular drawn from A to the hypotenuse : if the area of the triangle
ABC is 54 square inches and AB is 1 foot, shew that the area of the
triangle ADU is 19*44 square inches.

362 euclid's elements.

Proposition 20. Theorem.

Similar polygons may he divided into the same number of
similar triangles, having the same ratio each to each that the
polygons have ; and the polygons are to one another in the dupli-
cate ratio of their homologous sides.

D C K H

Let ABODE, FGHKL be similar polygons, and let AB and
FG be homologous sides.

Then (i) the polygons may be divided into the same number of
similar triangles ;
(ii) these triangles shall have each to each the same ratio

that the polygons have ;
(iii) the polygm ABODE slmll be to the polygon FGHKL in
the duplicate ratio of AB to FG.

Join EB, EO, LG, LH.

(i) Then because the polygon ABODE is similar to the
polygon FGHKL, Hyp.

.'. the L EAB = the l LFG,
and EA : AB :: LF : FG ; VI. Def 2.

.-. the A EAB is similar to the A LFG ; VI. 6.

.-. the ^ ABE = the z. FGL.
But because the polygons are similar, Hyp.

.'. the z- ABO = the l. FGH ; VL Def. 2.

.-. the remaining z. EBO = the remaining z. LGH.

And because the A" EAB, LFG are similar, Proved.
.-. EB : BA :: LG : GF;
and because the polygons are similar, Hyp.

.-. AB : BO :: FG : GH ; VL Def. 2.

.*. , ex cequali, EB : BO :: LG : GH ; V. 14.

that is, the sides about the equal z." EBO, LGH are
proportionals ;

.-. the A EBO is similar to the A LGH. VL 6.

BOOK VI. PROP. 20. 363

In the same way it may be proved that the A ECD is
similar to the A LHK.

.•. the polygons have been divided into the same number
of similar triangles.

(ii) Again, because the A EAB is similar to the A LFG,
.-. the A EAB is to the A LFG in the duplicate ratio
ofEB:LG; VI. 19.

and, in like manner,

the A EBC is to the A LGH in the duplicate ratio
of EB to LG ;
/. the A EAB : the A LFG :: the A EBC : the A LGH. V. 1.
In like manner it can be shewn that
the A EBC : the A LGH :: the A ECD : the A LHK ;
.-. the A EAB : the A LFG :: the A EBC : the A LGH

:: the A ECD : the A LHK.
But in a series of equal ratios, as each antecedent is to its
consequent so is the sum of the antecedents to the sum of
the consequents; [Addendo. v. 12.]

.-. the A EAB : the A LFG : : the fig. ABCDE : the fig. FGHKL.

(iii) Now the A EAB : the A LFG in the duplicate ratio

of AB : FG, VI. 19.

and the A EAB : the A LFG : : the fig. ABCDE : the fig. FGHKL;

.'. the fig. ABCDE : the fig. FGHKL in the duplicate ratio

of AB : FG. Q.E.D.

Corollary 1. Let a third proportional X be taken to
AB and FG,

then AB is to X in the duplicate ratio of AB : FG ;
but the fig. ABCDE : the fig. FGHKL in the duplicate
ratio of AB : FG ; Proved,

.-. AB : X :: the fig. ABCDE : the fig. FGHKL.
Hencey if three straight lines are proportionals, as the first is
to the third, so is any rectilineal figure described on the first to a
similar and similarly described rectilineal figure on the second.

Corollary 2. It follows that similar rectilineal figures
are to one another as the squares on their homologous sides. For
squares are similar figures and therefore are to one another
in the duplicata ratio of their sides.

364

EUCLID'S ELEMENTS.

Obs. The following theorem, taken from Euclid's Twelfth
Book, is given here as an important application of the pre-
ceding proposition.

Book XII. Proposition 1.

The areas of similar polygons inscribed in circles are to one another
as the squares on the diameters.

VI. Def, 2.
VI. 6.

III. 21.

Let ABODE and FGHKL be two similar polygons, inscribed in
the circles ACE, FHL, of which AM, FN are diameters.

Then shaM

the fig. ABODE : the fig. FGHKL :: the sq. on AM : the sq. on FN.

Join BM, AC and ON, FH.

Then since the polygon ABODE is similar to the polygon FGHKL,

.-. the L ABC = the L FGH,

and^ AB : BO : : FG : GH ;

.-. the A ABO is similar to the A FGH ;

.-. the L AOB = the L FHG.

But the L AOB = the Z- AMB ;

and the L FHG = the L FNG ;

.-. the L AMB = the L FNG.

Also in the A" ABM, FGN, the Z-« ABM, FGN are equal, being
rt. angles ; iii. 31.

hence the remaining Z.« BAM, GFN are equal ; i. 32.

and the A« ABM, FGN are similar : vi. 4.

.-. AB: FG :: AM : FN.

But the fig. ABODE : the fig. FGHKL in the duplicate ratio of
AB : FG, VI. 20.

that is, in the duplicate ratio of AM : FN. v. 16.

Hence

the fig. ABODE : the fig. FGHKL : : the sq. on AM : the sq. on FN.

VI. 20, Cor. 2.

BOOK VI. EXERCISES. 365

Obs. The following theorem, which forms Proposition 3
of Euclid's Twelfth Book, may be derived as a corollary
from the preceding proof.

Corollary. The areas of circles are to one another as the squares
on their diameters.

It has been shewn that
the fig. ABODE : the fig. FGHKL : : the sq. on AM : the sq. on FN :
and this is true however many sides the two polygons may have.

Suppose the polygons are regular ; then by sufficiently increasing
the number of their sides, we may make their areas dififer from the
areas of their circumscribed circles by quantities smaller than any
that can be named ; hence ultimately,

the ACE : the O FHL : : the sq. on AM : the sq. on FN.

EXERCISES ON PROPOSITIONS 19, 20.

1. If ABC is a triangle right-angled at A, and AD is drawn per-
pendicular to BC, shew that

(i) CB : BD in the duplicate ratio of CB to BA ;

(ii) The square on CB : the square on BA : : CB : BD ;

(iii) The A ABD : the A CAD in the duplicate ratio of BA
to AC.

2. In any triangle ABC, the sides AB, AC are cut by a line XY
drawn parallel to BC. If AX is one-third of AB, what part is the
triangle AXY of the triangle ABC ?

3. A trapezium ABCD has its sides AB, CD parallel, and its
diagonals intersect at O. If AB is double of CD, find the ratio of
the triangle AOB to the triangle COD.

4. ABC and XYZ are two similar triangles whose areas are
respectively 245 and 5 square inches. If AB is 21 inches in length,
find XY.

5. Shew how to draw a straight line XY parallel to the base
BC of a triangle ABC, so that the area of the triangle AXY may be
four-ninths of the triangle ABC.

6. Two oircles intersect at A and B, and at A tangents are
drawn, one to each circle, meeting the circumferences at C and D.
If AB, CB and BD are joined, shew that

the A CBA : the A ABD : : CB : BD.

366 EUCLID'S ELEMENTS.

Proposition 21. Theorem.

Rectilineal figures which are similar to the same rectilineal
figure^ are also similar to each other.

£12 £ll

Let each of the rectilineal figures A and B be similar to C.

Then shall A he similar to B.

For because A is similar to C, Hyp,

. • . A is equiangular to C,
and the sides about their equal angles are proportionals.

VI. Def. 2.

Again, because B is similar to C, Hyp,

.*. B is equiangular to C,
and the sides about their equal angles are proportionals.

VI. Def. 2.
.*. A and B are each of them equiangular to C, and have
their sides about the equal angles proportional to the corre-
sponding sides of C ;

.*. A is equiangular to B, Ax. 1.

and the sides of A and B about their equal angles are pro-
portionals; V. 1.

.'. A is similar to B.

Q.E.D.

book vi. prop. 22. 367

Proposition 22. Theorem.

If four straight lines he proportional and a pair of similar
rectilineal figures be similarly desciihed on the first and second^
and also a pair on the third and fourth, these figures shall be
proportional.

Conversely, if a rectilineal figure on the first of four straight
lines be to the similar and similarly desciibed figure on the
second as a reetilineal figure on the third is to the similar and
similarly described figure on the fourth, the four straight lines
shall be proportional.

F G H

First. Let AB, CD, EF, GH be proportionals,

so that AB : CD :: EF : GH ;
and let similar figures KAB, LCD be similarly described on
AB, CD, and also let similar figures MF, NH be similarly
described on EF, GH.

Then shall

the fig. KAB : the fig. LCD :: the fig. MF : the fig. NH.

To AB and CD take a third proportional X ; VI. 11.
and to EF and GH take a third proportional O ;

then AB : CD :: CD : X, Constr.

and EF : GH :: GH : O.

But AB : CD :: EF : GH ; Hyp.

.-. CD : X :: GH : O, V.' 1.

.*., ex cequali, AB : X :: EF : O. V. 14.

But AB : X : : the fig. KAB : the fig. LCD ; VI. 20, Cw.
and EF : O :: the fig. MF : the fig. NH ;
.-. the fig. KAB : the fig. LCD :: the fig. MF : the fig. NH.

V. 1

H.S.B. 2a

368 EUCLID'S ELEMENTS.

Conversely,
Let the fig. KAB : the fig. LCD :: the fig. MF ; the fig. NH.

Tlien shall AB : CD :: EF : GH.

To AB, CD, and EF take b. fourth proportional PR : vi. 12.
and on PR describe the fig. SR similar and similarly situated
to either of the figs. MF, NH. VL 18.

Then because AB : CD :: EF : PR, Const,

.-., by the former part of the proposition,
the fig. KAB : the fig. LCD :: the fig. MF : the fig. SR.
But, by hypothesis,

the fig. KAB : the fig. LCD :: the fig. MF : the fig. NH ;
.-. the fig. MF : the fig. SR :: the fig. MF : the fig. NH, V. 1.

.-. the fig. SR = thefig. NH.
And since the figs. SR and NH are similar and similarly
situated, Comtr,

.'. PR = GH*
Now AB : CD :: EF : PR; Constr,

.', AB : CD :: EF : GH.

Q.E.D.

* Euclid here assumes that if two similar and similarly situaied
figures are equate their homologous sides are equal. The proof is
easy and may be left as an exercise for the student.

Definition. AVhen there are any number of magnitudes
of the same kind, the first is said to have to the last the
ratio compounded of the ratios of the first to the second^ of
the second to the third, and so on up to the ratio of the
last hut one to the last magnitude. [Book v. Def. 11.]

BOOK VI. PROP. 23.

369

Proposition 23. Theorem.

Parallelograms which are equiangular to one another have to
one another the ratio which is compounded of the ratios of their
sides.

A D H

^ :

K L M E F

Let the par"* AC be equiangular to the par" CF, having the
L BCD equal to the l ECG.

Then shall the par^ AC Imve to the par'^ CF the ratio com-
pounded of the ratios BC : CG and DC : CE.

Let the par"* be placed so that BC and CG are in a st. line;
then DC and CE are also in a st. line. i. 14.

Complete the par" DG.

Take any st. line K,

and to BC, CG, and K find a fourth proportional L ; VI. 1 2.

and to DC, CE, and L take a. fourth proportional M ;

then BC : CG : : K : L,
and DC : CE :: L : M.
But K : M is the ratio compounded of the ratios

K : Land L : M; V. Def 11,

that is, K : M is the ratio compounded of the ratios

BC : CG and DC : CE.

Now the par™ AC : the par™ DG

and the par™ DG : the par™ CF

BC : CG
K : L,
DC : CE
L : M;
K : M.

VI. 1.
Constr.

VI. 1.
Constr.
V. 14.

.'., ex ceqy^ali, the par™ AC : the par™ CF

But K : M is the ratio compounded of the ratios of the sides ;

.•. the par™ AC has to the par™ CF the ratio compounded

of the ratios of the sides. Q.E.D.

Exercise. The areas of two triangles or parallelograms are to
one another in the ratio compounded of the ratios of their bases and
of their altitudes,

370 Euclid's elements.

Proposition 24. Theorem.

Parallelograms about a diagonal of any parallelogram are
similar to the whole parallelogram and to one another,

A E B

D K C

Let ABCD be a par™ of which AC is a diagonal ;
and let EG, HK be par"* about AC.

Then shall the par^ EG, HY^he similar to the par"^ ABCD, and
to one another.

For because DC is par^ to GF,

.-. thez- ADC = the .1 AGF; I 29.

and because BC is par* to EF,
.-. the ^ ABC = the l AEF ; L 29.

and each of the z." BCD, EFG is equal to the opp. l BAD,

.-. the L BCD = the l EFG; L 34.

.*. the par™ ABCD is equiangular to the par™ AEFG.

Again in the A' BAC, EAF,
because the l ABC = the l AEF, I. 29.

and the l BAC is common ;
.*. the remaining l BCA = the remaining z. EFA; L 32.
.*. the A' BAC. EAF are equiangular to one another ;

.*. AB : BC :: AE : EF. VI. 4.

But BC = AD, and EF = AG ; L 34.

.-. AB : AD :: AE : AG.
Similarly DC : CB :: GF : FE,
and CD : DA : : FG : GA ;
.*. the sides of the par"* ABCD, AEFG about their equal

angles are proportional ;
.-. the par™ ABCD is similar to the par™ AEFG. VL Def. 2.

In the same way the par™ ABCD may be proved similar
to the par™ FHCK,

.-. each of the par™" EG, HK is similar to the whole par™;
/. the par™ EG is similar to the par™ HK. vi. 21.

Q.£.Dt

BOOK VI. PROP. 25. 371

Proposition 25. Problem.

To describe a rectilineal figure which shall he eqml to one and
similar to another rectilineal figure.

Let E and S be the two given rectilineal figures.

It is required to describe a figure equal to the fig, E and similar
to the fig. S.

On AB a side of the fig. S describe a par" ABCD equal to S ;
and on BC describe a par" CBGF equal to the fig. E, and
having the l CBG equal to the z. DAB ; I. 45.

then AB and BG are in one st. line, and also DC and CF in
one St. line.

Between AB and BG find a mean proportional HK ; vi. 13.
and on HK describe the fig. P, similar and similarly situated
to the fig. S. VI. 18.

Then P shall be the figure required.

Because AB : HK :: HK : BG, Constr.

.'. AB : BG :: the fig. S : the fig. P. VI. 20, Cor.

But AB : BG :: the par" AC : the par" BF ; vi. 1.
.-. the fig. S : the fig. P :: the par" AC : the par" BF; V. 1.

and the fig. S = the par" AC ; Cmistr.

.-. the fig. P = the par" BF

= the fig. E. Constr.

And since, by construction, the fig. P is similar to the fig. S,

.*. P is the figure required.

Q.E.F.

372 Euclid's elements.

Proposition 26. Theorem.

If two similar parallelograms have a common angle, and are
similarly situated, they are db&ui the same diagonal.

Let the par"" ABCD, AEFG be similar and similarly situated,
and have the common angle BAD.

Then shall the par^ ABCD, AEFG be about the same diagonal.

Join AC.
Then if AC does not pass through F, if possible let it cut
FG, or FG produced, at H.

Through H draw HK par^ to AD or BC. I. 31.

Then the par™' BD and KG are similar, since they are about
the same diagonal AHC ; VI. 24.

.-. DA : AB :: GA : AK.
But because the par""* BD and EG are similar ; Hyp.
.'. DA : AB :: GA : AE ; VL Def. 2.

.-. GA : AK :: GA : AE;

. '. AK = AE, which is impossible ;
.'. AC must pass through F ;
that is, the par"' BD, EG are about the same diagonal.

Q.E.D.

BOOK VI. PROP. 30. 373

Ohs. Propositions 27, 28, 29 being cumbrous in form and of little
value as geometrical results are now very generally omitted.

Definition. A straight line is said to be divided in
extreme and mean ratio, when the whole is to the greater
segment as the greater segment is to the less.

[Book VI. Def. 5.J

Proposition 30. Problem.
To divide a given straight liiie in extreme and mean ratio.

C B

Let AB be the given st. line.
It is required to divide AB in extreme and mean ratio.

Divide AB in C so that the rect. AB, BC may be equal to
the sq. on AC. li. 11.

Then because the rect. AB, BC = the sq. on AC,

.-. AB : AC :: AC : BC. VI. 17.

Q.E.F.

EXERCISES.

1. ABCDE is a regular pentagon ; if the lines BE and AD
intersect in O, shew that each of them is divided in extreme and
mean ratio.

2. If the radius of a circle is cut in extreme and mean ratio,
the greater segment is equal to the side of a regular decagon inscribed
in the circle.

374

Euclid's elements.

Proposition 31. Theorem.

In a right-angled triangle, any rectilineal figure described
on the hypotenuse is equal to the sum of the two similar and
similarly described figures on the sides containing the right angle.

Let ABC be a right-angled triangle of which BC is the
hypotenuse; and let P, 'Q, R be similar and similarly
described figures on BC, CA, AB respectively.

Then shall the fig, P be equal to the sum of the figs, Q and R.

Draw AD perp. to BC.

VI. 8.

Then the A" CBA, ABD are similar ;
.-. CB : BA :: BA : BD;

.-. CB : BD : : the fig. P : the fig. R ; VL 20, Car.
.*., inversely, BD : BC :: the fig. R : the fig. P. v. 2.

In like manner, DC : BC : : the fig. Q : the fig. P ;
.-. the sum of BD, DC : BC :: the sum of figs. R, Q : fig. P;

V. 15.
but BC = the sum of BD, DC ;
.'. the fig. P = the sum of the figs. R and Q.

Q.E.D.

Note. This proposition is a generalization of Book I. , Prop. 47.
It will be a useful exercise for the student to deduce the general
theorem (vi. 31) from the particular case (i. 47) with the aid of
VI. 20, Cor. 2.

EXERCISES ON PROPS. 19-31. 375

EXERCISES.

1. In a right-angled triangle if a perpendicular is drawn from
the right angle to the opposite side, the segments of the hypotenuse
are in the duplicate ratio of the sides containing the right angle.

2. If, in Proposition 31, the figure on the hypotenuse is equal
to the given triangle, the figures on the other two sides are respec-
tively equal to the parts into which the triangle is divided by the
perpendicular from the right angle to the hypotenuse.

3. AX and BY are medians of the triangle ABC which meet in
Q : if XY is joined, compare the areas of the triangles AGB, XGY.

4. Shew thcU similar triaruflea are to one another in the duplicate
raiio of (i) corresponding medians, (ii) the radii of their inscribed
circles, (iii) the radii of their circumscribed circles.

5. DEF is the pedal triangle of the triangle ABC ; prove that
the triangle ABC is to the triangle DBF in the duplicate ratio of
AB to BD. Hence shew that

the fig. AFDC : the A BFD : : AD^ : BD^.

6. The base BC of a triangle ABC is produced to a point D
such that BD : DC in the duplicate ratio of BA : AC. Shew that
AD is a mean proportional between BD and DC.

7. Bisect a triangle by a line drawn parallel to one of its sides.

8. Shew how to draw a line parallel to the base of a triancle so
as to form with the other two sides produced a triangle douole of
the given triangle.

9. If through any point within a triangle lines are drawn from
the angles to cut the opposite sides, the segments of any one side
will have to each other the ratio compounded of the ratios of the
segments of the other sides.

10. Draw a straight line parallel to the base of an isosceles
triangle so as to cut off a triangle which has to the whole triangle
the ratio of the base to a side.

11. Through a given point, between two straight lines con-
taining a given angle, draw a line which shall cut off a triangle
equal to a given rectilineal figure.

Obs. The 32nd Proposition as given by Euclid is de-
fective, and as it is never applied, we have omitted it.

376

Euclid's elements.

Proposition 33. Theorem.

In equal circles, angles, whether at the centres or the dr-
citmferences, have the same ratio as the arcs on which they stand:
so also have the sectors.

Let ABC and DEF be equal circles, and let BGC, EHF be
angles at the centres, and BAG and EDF angles at the O**^.

Then shall

(i) the L BGC : the l EHF :: the arc BC : the arc EF ;
(ii) the L BAC : the l EDF :: the arc BC : the arc EF ;
(iii) the sector BGC : the sector EHF :: the arc BC : the
arc EF.

Along the O** of the O ABC take any number of arcs
CK, KL each equal to BC ;

and along the C* of the DEF take any number of arcs
FM, MN, NR each equal to EF.

Join GK, GL, HM, HN, HR.

(i) Then the l' BGC, CGK, KGL are all equal,
for they stand on the equal arcs BC, CK, KL : lii. 27.
.-. the L BGL is the same multiple of the l BGC that the
arc BL is of the arc BC.
Similarly, the z_ EHR is the same multiple of the z_ EHF
that the arc ER is of the arc EF.

And if the arc BL = the arc ER,

the L BGL = the l EHR ; m. 27.

and if the arc BL is greater than the arc ER,

the /- BGL is greater thati the l EHR ;

and if the arc BL is less than the arc ER,

the L BGL is less than the l EHR.

BOOK VI. PROP. 33.

377

Now since there are four magnitudes, namely the
iJ BGC, EHF and the arcs BC, EF; and of the antecedents
any equimultiples have been taken, namely the z. BGL and
the arc BL ; and of the consequents any equimultiples have
been taken, namely the ^ EHR and the arc ER :
and since it has been proved that the l BGL is greater than,
equal to, or less than the l. EHR, according as BL is greater
than, equal to, or less than ER ;

.*. the four original magnitudes are proportionals ; V. Bef, 5.
that is, the l. BGC : the l EHF :: the arc BC : the arc EF.

(ii) And since the l BGC = twice the l BAC, hi. 20.

and the l EHF = twice the l EDF;
.*. the L BAC : the l EDF :: the arc BC : the arc EF. V. 8.

(iii) Join BC, CK; and in the arcs BC, CK take any
points X, O.

Join BX, XC, CO, OK.
Then in the A" BGC, CGK,
( BG == CG,

Because-^ GC = GK,

[and the l BGC = the l CGK ;

. * . BC = C K j

and the A BGC = the A CGK.
And because the arc BC = the arc CK,
.*. the remaining arc BAC = the remaining arc CAK :

.-. the L BXC = the l COK; III. 27.

.*. the segment BXC is similar to the segment COK ;

III. Bef. 10.
and these segments stand on equal chords BC, CK ;

.*. the segment BXC = the segment COK. III. 24.
And the A BGC = the A CGK;
.'. the sector BGC = the sector CGK.

III. 27.
I. 4.

Constr,

378

EUCLID S ELEMENTS.

Similarly it may be shewn that the sectors BGC, CGK,
KGL are all equal ;
and likewise the sectors EHF, FHM, MHN, NHR are all equal.

.'. the sector BGL is the same multiple of the sector BGC

that the arc BL is of the arc BC ;
and the sector EHR is the same multiple of the sector EHF

that the arc ER is of the arc EF.

And if the arc BL = the arc ER,

the sector BGL = the sector EHR : Proved,

and if the arc BL is greater than the arc ER,

the sector BGL is greater than the sector EHR :

and if the arc BL is less than the arc ER,

the sector BGL is less than the sector EHR.

Now since there are four magnitudes, namely, the sectors
BGC, EHF and the arcs BC, EF; and of the antecedents any
equimultiples have been taken, namely the sector BGL and
the arc BL ; and of the consequents any equimultiples have
been taken, namely the sector EHR and the arc ER :
and since it has been shewn that the sector BGL is greater
than, equal to, or less than the sector EHR, according as the
arc BL is greater than, equal to, or less than the arc ER ;
.*. the four original magnitudes are proportionals ;

V. Def. 5.
that is,
the sector BGC : the sector EHF :: the arc BC : the arc EF.

Q.E.D.

BOOK VI. QUESTIONS FOR REVISION. 379

QUESTIONS FOR REVISION.

1. Explain why the operation known as Alternately requires
that the four terms of a proportion should be of the same kind.
Show that this is unnecessary in the case of Inversely.

2. State and prove algebraically the theorem known as Com-
ponendo. In what proposition is this principle applied ?

3. Enunciate and prove algebraically the operation used in
Book VI. under the name Ex JEquali.

Also prove the same theorem in the following more general form :

If there are two sets of magnitudes , such that the first is to the
second of the first set as the first to the second of the other set, and
the second to the third of the first set as the second to the third
of the other, and so on throughout : then the first shall be to the last
of the first set as the first to the last of the other.

4. Explain the operation Addendo, and give an algebraical
proof of it. In what proposition of Book vi. is this operation
employed ?

5. Give the geometrical and algebraical definitions of the ratio
compounded of given ratios, and shew that the two definitions agree.

By what artifice would Euclid represent the ratio compounded
of the ratios A : B and G i Dt

6. Two parallelograms ABCD, EFGH are equiangular to one
another : if AB, BC are respectively 21 and 18 inches in length, and
if EF, FG are 27 and 35 inches ; shew that the areas of the parallelo-
grams are in the ratio 2 : 5.

7. liA'.B=X'.Y, &nd C : B = Z: T;
shew that A + G : B = X + Z: T.

In what proposition of Book vi. is this principle used ?

Explain and illustrate the necessity of the step invertendo in this
proposition.

8. When is a straight line said to be divided in extreme and
m^ean ratio ?

If a line 10 inches in length is so divided, shew that the lengths
of the segments are approximately 6*2 inches and 3*8 inches.

Shew also that the segments of any line divided in extreme and
mean ratio are incommensurable.

380 Euclid's elements.

Proposition B. Theorem.

If the vertical angle of a triangle he bisected by a straight
line which cuts the base, the rectangle contained by the sides of
the triangle shall be equal to the rectangle contained by the
segments of the base^ together with the square on the straight line
which bisects the angle.

E
Let ABC be a triangle, having the l BAG bisected by AD.

Then shall

the rect. BA, AC = the rect. BD, DC, with the sq. on AD.

Describe a circle about the A ABC, iv. 5.

and produce AD to meet the O** in E.

Join EC.

Then in the A* BAD, EAC,
because the l BAD = the l EAC, Hyp.

and the z. ABD = the z_ AEC in the same segment ; IIL 21.
.*. the remaining l BDA = the remaining l ECA ; i. 32.
that is, the A BAD is equiangular to the A EAC.

.-. BA : AD :: EA : AC; VL 4.

.-. the rect. BA, AC = the rect. EA, AD, VI. 16.

= the rect. ED, DA, with the sq. on AD.

11. 3.

But the rect. ED, DA = the rect. BD, DC; in. 35.

.*. the rect. BA, AC = the rect. BD, DC, with the sq. on AD.

Q.E.D.

exercise.

If the vertical angle BAC is externally bisected by a straight
line which meets the base in D, shew that the rectangle contained
by BA, AC together with the square on AD is equal to the rectangle
contained by the segments of the bas^,

BOOK VI. PROP. C. 381

Proposition C. Theorem.

If from the vertical angle of a triangle a straight line be
drawn perpendicular to the base, the rectangle contained by the
sides of the triangle shall be equal to the rectangle contained
by the perpendicular and the diameter of the circle described
about the triangle.

Let ABC be a triangle, and let AD be the perp. from A
to the base BC.

Then the red. BA, AC shall be equal to the rectangle contained
by AD and the diameter of the circle circumscribed about the
A ABC.

Describe a circle about the A ABC ; iv. 5.

draw the diameter AE, and join EC.

Then in the A' BAD, EAC,
the rt. angle BDA = the rt. angle EGA, in the semicircle ECA,
and the l ABD = the l AEC, in the same segment ; iii. 21.
.*. the remaining l BAD = the remaining l EAC ; i. 32.
that is, the A BAD is equiangular to the A EAC ;

.*. BA : AD :: EA : AC; VI. 4.

.•. the rect. BA, AC = the rect. EA, AD. VI. 16.

Q.E.D.

382 euclid's elements.

Proposition D. Theorem.

The rectangle contained by the diagonals of a quadrilateral
inscribed in a circle is equal to the sum of the two rectangles
contained by its opposite sides.

Let ABCD be a quadrilateral inscribed in a circle, and
let AC, BD be its diagonals.

Then the reel. AC, BD shall be equal to the sum of the rect-
angles AB, CD and BC, AD.

Make the l DAE equal to the l BAC ; I. 23.

to each add the l EAC,
then the l DAC = the l EAB.

Then in the A" EAB, DAC,
the L EAB = the l DAC,
and the l ABE = the l ACD in the same segment; ill. 21.
.'. the A* EAB, DAC are equiangular to one another ; L 32.

.*. AB : BE :: AC : CD; VL 4.

.-. the rect. AB, CD = the rect. AC, Ea VI. 16.

Again in the A' DAE, CAB,
the L DAE = the l CAB, Constr,

and the l ADE = the l ACB, in the same segment, III. 21.
.•. the A" DAE, CAB are equiangular to one another; I. 32.

.*. AD : DE :: AC : CB; VL 4.

.-. the rect. BC, AD = the rect. AC, DE. VL 16.

But the rect. AB, CD = the rect. AC, EB. Proved,

.'. the sum of the rects. BC, AD and AB, CD = the sum of

the rects. AC, DE and AC, EB ;
that is, the sum of the rects. BC, AD and AB, CD

==the rect. AC, BD. ll. 1.

Q.E.D.

EXERCISES ON PROPS. B, C, D. 383

Note. Propositions B, C, and D do not occur in Euclid, but
were added by Robert Simson,- who edited Euclid's text in 1756.

Prop. D is usually known as Ptolemy's theorem, and it is the
particular case of the following more general theorem :

The rectangle contained by the diagoncds of a quadrilateral is less
than the sum of the rectangles contained by its opposite sides, unless a
circle can be circumscribed about the quadrilateral, in which case it is
eqtial to that sum.

EXERCISES.

1. ABC is an isosceles triangle, and on the base, or base pro-
duced, any point X is taken : shew that the circumscribed circles of
the triangles ABX, ACX are equal.

2. From the extremities B, C of the base of an isosceles triangle
ABC, straight lines are drawn perpendicular to AB, AC respectively,
and intersecting at D : shew that the rectangle BC, AD is double of
the rectangle AB, DB.

3. If the diagonals of a quadrilateral inscribed in a circle are at
right angles, the sum of the rectangles contained by the opposite
sides is double the area of the figure.

4. ABCD is a quadrilateral inscribed in a circle, and the diagonal
BD bisects AC; shew that the rectangle AD, AB is equal to the rect-
angle DC, CB.

6. If the vertex A of a triangle ABC is joined to any point in
the base, it will divide the triangle into two triangles such that their
circumscribed circles have radii in the ratio of AB to AC.

6. Construct a triangle, having given the base, the vertical
angle, and the rectangle contained by the sides.

7. Two triangles of equal area are inscribed in the same circle :
shew that the rectangle contained by any two sides of the one is to
the rectangle contained by any two sides of the other as the base of
the second is to the base of the first.

8. A circle is described round an equilateral triangle, and from
any point in the circumference straight lines are drawn to the
angular points of the triangle : shew that one of these straight lines
is equal to the sum of the other two.

9. ABCD is a quadrilateral inscribed in a circle, and BD bisects
the angle ABC : if the points A and C are fixed on the circumference
of the circle and B is variable in position, shew that the sum of AB
and BC has a constant ratio to BD.

II. a. E. 2 r.

THEOREMS AND EXAMPLES ON BOOK VL

I. ON HARMONIC SECTION.

1. To divide a given straight line internally and extemaUy so that
its segments may he in a given ra4,io.

9 /

L M A P\ ye Q

Let AB be the given st. line, and L, M two other st. lines which
determine the given ratio.

It is required to divide AB internally and externally in the ratio L : M.

Through A and B draw any two par^ st. lines AH, BK.
From AH cut otf Aa equal to L,
and from BK cut oflf B?> and Bb' each equal to M, B6' being taken in
the same direction as Aa, and B6 in the opposite direction.

Join a&, cutting AB in P ;
join ab', and produce it to cut AB externally at Q.

Then AB shall he divided internally at P and externally al Q,
80 that AP : PB = L : M.

and AQ : QB = L : M.

The proof follows at once from Euclid vi. 4.

Note. The solution is singular ; that is, only one internal and one
external point can he found that will divide the given straight line
into segments which have the given' ratio.

HARMONIC SECTION. 385

Definition. A finite straight line is said to be cut
harmonically when it is divided internally and externally
into segments which have the same ratio.

A P B Q

Thus AB is divided harmonically at P and Q, if

AP : PB = AQ : QB.
P and Q are said to be harmonic conjugates of A and B.

Now by taking the above proportion cUtemcUely,
we have PA : AQ= PB : BQ ;

from which it is seen that if P and Q divide AB internally and
externally in the same ratio, then A and B divide PQ internally and
externally in the same ratio ; hence A and B are harmonic conjugates
of P and Q.

Example. The hose of a triangle is divided harmonically by the
internal and external bisectors of the vertical angle :
for in each case the segments of the base are in the ratio of the other
sides of the triangle. [Euclid vi. 3 and A.]

Obs. We shall use the terms A rithmetic^ Geometric, and Harmonic
Means in their ordinary Algebraical sense.

1. If AB is divided internally at P and externally at Q in the
same ratio, then AB is the harmonic mean between AQ and AP.

For, by hypothesis, AQ : QB = AP : PB ;
.-. , alternately, AQ : AP = QB : PB,

that is, AQ : AP = AQ - AB : AB - AP ;

.*. AP, AB, AQ are in Harmonic Progression.

2. JfAB is divided harmonically at P and Q, and O is the middle
point of AB ;

then OP. 0Q-=0A2.

A OPS Q

For since AB is divided harmonically at P and Q,
.-. AP : PB = AQ : QB ;
/. AP-PB : AP+PB = AQ-QB :AQ + QB,
or, 20P : 20A=20A : 20Q ;

.-. 0P.0Q=0A2.

Conversely, if OP . OQ = OA^ ,

it may be shewn that

AP:PB = AQ:QB;
that is, that AB is divided harmonically at P and Q.

386

EUCLID'S ELEMENTS.

3. The Arithmetic, Geometric and Harmonic means ofUoo straight
lilies may he thiis represented grajihicaily.

In the adjoining figure, two tan- K

gents AH, AK are drawn from any
external point A to the circle PHQK ;
HK is the chord of contact, and the
St. line joining A to the centre O cuts
the 0<* at P and Q.

Then (i) AO is the Arithmetic
mean between AP and AQ : for clearly
AO = MAP + AQ).

(ii) AH is the Geometric mean between AP and AQ :

for AH2=AP . Aa m. 36.

(iii) AB is the Harmonic mean between AP and AQ :

for OA . 0B = 0P2 ; Ex. 1, p. 251.

.*. AB is cut harmonically at P and Q. Ex. 2, p. 385.

That is, AB is the Harmonic mean between AP and AQ.-

And from the similar triangles OAH, HAB,

OA: AH=AH : AB,
.-. A0.AB = AH2; VI. 17.

/. the Geometric mean between two straight lines is the mean propor-
tional between their Arithmetic and Harmonic means.

4. Given the base of a. triangle and the ratio of the other sides, to
find the locus of the vertex.

Let BC be the given base, and let
BAG be any triangle standing upon A^

it, such that BA : AC = the given
ratio.

It is required to find the locus of A.

Bisect the L BAG internally and ^
externally by AP, AQ.

Then BG is divided internally at P, and externally at Q,

so that BP : PG=BQ : QG = the given ratio ;

.'. P and Q are fixed points.

And since AP, AQ are the internal and external bisectors of the
Z-BAG,

.'. the L PAQ is a rt. angle ;
.*. the locus of A is a circle described on PQ as diameter.

ExKRCiSE. Given three points B, P, G in a straight line : find the
locus of points at which BP and PG subtend equal angles.

HARMONIC SECTION. 387

DEFINITIONS.

1. A series of points in a straight line is called a range.
If the range consists of four points, of which one pair are
harmonic conjugates with respect to the other pair, it is
said to be a harmonic range.

2. A series of straight lines drawn through a point is
called a pencil.

The point of concurrence is called the vertex of the
pencil, and each of the straight lines is called a ray.

A pencil of four rays drawn from any point to a har-
monic range is said to be a harmonic pencil.

3. A straight line drawn to cut a system of lines is
called a transversal.

4. A system of four straight lines, no three of which
are concurrent, is called a complete quadrilateral.

These straight lines will intersect two and two in six
points, called the vertices of the quadrilateral; the three
straight lines which join the opposite vertices are diagonals.

Theorems on Harmonic Section.

1. If a transversal is draion parallel to one ray oj a karTUonic
pencil, the other three rays intercept equal parts upon it: and con-
versely.

2. Any traniwersal is cuttharmonically by the rays of a harmonic
pencil,

3. In a harmonic pencil, if owe ray bisect the angle between the
other pair of rays, it is perj^endicular to its conjugate ray. Conversely,
if one pair of rays form a right angle, then they bisect internally and
externally the angle between the other pair.

4. If A, P, B, Q and a, p, b, q are harmonic ranges, one on each
of two given straight lines, and if Aa, Pp, Bb, the straight lines which
join three pairs of coj-responding points, meet at S ; then will Qq also
pa^ through S.

5. If tioo straight lines intersect at A, and if A, P, B, Q and
A, p, b, q are two harmonic ranges one on each straight line {the points
corresponding as indicated by the letters), then Pp, Bb, Qq will be con-
current : also Pq, Bb, Qp tvill be concurrent.

6. Use Theorem 5 to prove that in a complete quadrilateral in
which the three diagonals are drawn, the straight line joining any pair
of opposite vertices is cvJt harmonically by the other two diagonals.

388 EUCLID'S ELEMENTS.

IL ON CENTRES OF SIMILARITY AND SIMILITUDE.

1. If any two unequal similar figures are placed so that their
homologous sides are paraXlel^ the lines joining coi'responding points in
the two figures meet in a point, whose distances from any two corre-
sponding points are in the ratio of any pair of homologous sides.

Let ABCD, A'B'C'D' be two similar figures, and let them be placed
so that their homologous sides are parallel ; namely, AB, Bu, CD,
DA parallel to A'B', B'C, CD', D'A' respectively.

Then shall AA', BB', CC, DD' meet in a point, whose distances from
any two corresponding points shall be in the ratio of any pair of
homologous sides.

Let AA' meet BB', produced if necessary, in S.

Then because AB is par^ to A'B' ; ffyp*

.: the A" SAB, SA'B' are equiangular ;

.-. SA:SA' = AB :A'B'; vi. 4.

.*. AA' divides BB', externally or internally, in the ratio of AB to A'B'.

Similarly it may be shewn that CC divides BB' in the ratio of
BC to B'C.

But since the figures |ire similar,
BC:B'C' = AB:A'B';

.'. AA' and CC divide BB' in the same ratio :
that is, AA', BB', CC meet in the same point S.
In like manner it may be proved that DD' meets CC in the
point S.

.*. AA', BB', CC, DD' are concurrent, and each of these lines is
divided at S, externally or internally, in the ratio of a pair of
homologous sides of the two figures. Q.E.D.

Cor. If any line is dratvn through S meeting any pair of homolo-
gous sides in K and K', the ratio SK : SK' is constant , and equal to the
ratio of any pair of homologous sides.

Note. It will be seen that the lines joining corresponding points
are divided externally or internally at S according as the correspond-
ing sides are drawn in the same or in opposite directions. In either
case the point of concurrence S is called a centre of similarity of the
two figures.

CENTRES OF SIMILITUDE. 389

2. A common tangeiU STT' to two circles whose centres are C, C,
meets the line of centres in S. If through S a7iy straight line is dravm
meeting these two circles in P, Q, and P', Q', respectively, then the
i^dii CP, CQ shall he respectively parallel to C'P', C'Q'. Also the
rectangles SQ . SP', SP . SQ' shall each he equal to the rectangle
ST . ST'.

Join CT, CP, CQ and CT', CP', C'Q'.
Then since each of the Z." CTS, CT'S is a right angle, iii. 18.

/. CT is pari to CT' ;
.-. the A" SCT, SuT' are equiangular ;
/. SC : SC' = CT : CT'
= CP:C'P';
.-. the A« SCP, SC'P' are similar ; vi. 7.

/. the L SCP= the SC'P' ;

/. CP is pari to CP'.
Similarly CQ is par^ to CQ'.

Again, it easily follows that TP, TQ are par^ to T'P', T'Q'
respectively ;

.-. the A" STP, ST'P' are similar.
Now the rect. SP . SQ=the sq. on ST ; iii. 36.

.-. SP : ST = ST : SQ, vi. 16.

andSP: ST = SP' :ST';
/. ST :SQ = SP' :ST';
/. the rect. ST . ST'=SQ . SP'.
In the same way it may be proved that

the rect. SP . SQ' = the rect. ST . ST'.

Q.E.D.

Cob. 1. It has been proved that

SC:SC = CP:C'P';
thus the external common tangents to the two circles meet at a point
S which divides the line of centres externally in the ratio of the radii.

Similarly it may be shewn that the transverse common tangents
meet at a point S' which divides the line of centres internally in the
ratio of the radii.

Cob. 2. CC is divided harmonically at S and S'.

Definition. The points S and S' which divide externally and
internally the line of centres of two circles in the ratio of their radii
are called the eztemal and internal centres of similitude respectively.

390 Euclid's elements.

examples on centres of similitude.

1. Inscribe a square in a given triangle.

2. In a given triangle inscribe a triangle similar and similarly
situated to a given triangle.

3. Inscribe a square in a given sector of circle, so that two
angular points shall be on the arc of the sector and the other two
on the bounding radii.

4. In the figure on page 298, if DI meets the inscribed circle in
X, shew that A, X, D^ are collinear. Also if AK meets the base in
Y shew that W^ is divided harmonically at Y and A.

5. With the notation on pa^e 302 shew that O and G are respec-
tively the external and internal centres of similitude of tlie circum-
scribed and nine-points circle.

6. If a variable circle touches two fixed circles, the line joining
tlieir points of contact passes through a centre of similitude. Dis-
tingtiish between the difie^^ent cases.

7. Describe a circle which shaZl touch tvjo given circles and pass
through a given point.

8. Describe a circle which shall touch three given circles.

9. Cj, Cg, Cg are the centres of three given circles ; S\, Sj, are the
internal and external centres of similitude of the 'pair of circles whose
centres are 0^, Cg, and S'a, S2, S'g, S3, have similar m^eanings with
regard to the other two pairs of cirdes : shew that

(i) S'lCj, S'aCg, S'gCg are concurrent ;

(ii) the six points Sj, Sg, Sg, S'j, S'g, S'g, lie three and three on
four straight lines. [See Ex. 1 and 2, pp. 400, 401.]

III. ON POLE AND POLAR.
DEFINITIONS.

1. If in any straight line drawn from the centre of a
circle two points are taken such that the rectangle contained
by their distances from the centre is equal to the square on
the radius, each point is said to be the inverse of the other.

Thus in the figure given on the following page, if O is
the centre of the circle, and if OP . OQ = (radius)^, then each
of the points P and Q is the inverse of the other.

It is clear that if one of these points is within the circle
the other must be without it.

POLE AND POLAR.

391

2. The polar of a given point with respect to a given
circle is the straight line drawn through the inverse of the
given point at right angles to the line which joins the given
point to the centre : and with reference to the polar the
given point is called the pole.

Thus in the adjoining figure, if OP . OQ = (radius )2, and if through

P and Q, LM and HK are drawn perp. to OP ; then HK is the polai'
of the point P, and P is the pole of the st. line H K with respect to
the given circle : also LM is the polar of the point Q, and Q the
pole of LM.

It is clear that the polar of an external point must intersect the
circle, and that tlie polar of an interned point must fall without it :
also that the polar of a point on the circumference is the tangent at
that point.

1. Now it has been proved [see Ex. 1,
page 251] that if from an external point P
two tangents PH, PK are drawn to a circle,
of which O is the centre, then OP cuts the
chord of contact HK at right angles at Q,
so that

0P.0Q = (radius)2;
.*. H K is the polar of P with respect to the
circle. Def. 2.

Hence we conclude that

The polar of an external point loith
reference to a circle is the chord of contact
of tangents draxvn from the given point to the circle.

392

EUCLID'S ELEMENTS.

2. //* A and P are any two points ^ and if the polar of A toith
respect to any circle passes through P, then the polar of P mviM pass
through A.

Let BC be the polar of the point A
with respect to a circle whose centre is
O, and let BC pass through P.

Then shall the polar of P pass through A.

Join OP ; and from A draw AQ perp.
to OP. We shall shew that AQ is the
polar of P.

Now since BC is the polar of A,
.'. the L ABP is a rt. angle ;

Def. 2, page 391.

and the L AQP is a rt. angle : Gonstr.

.'. the four points A, B, P, Q are coney clic;

.*. OQ . 0P= OA . OB III. 36.

= (radius)^ for CB is the polar of A :
.*. P and Q are inverse points with respect to the given circle.

And since AQ is perp. to OP,
AQ is the polar of P.

That is, the polar of P passes through A.

Q.E.D.

Note. A similar proof applies to the case when the given point
A is without the circle, and the polar BC cuts it.

The above Theorem is known as the Reciprocal Property of Pole
and Polar.

3. To prove that the locus of the intersection of tangents dravm to
a circle at the extremities of all chords which pass through a given
point within the circle is the polar of that point.

Let A be the given point within the
circle. Let HK be any chord passing
through A ; and let the tangents at H
and K intersect at P.

It is required to prove that the locus of
P is the polar of the point A.

I. To shew that P lies on the polar
of A.

Since H K is the chord of contact of
tangents drawn from P,

.-. HK is the polar of P. Ex. 1, p. 391.

POLE AND POLAR. 393

But H K, the polar of P, passes through A ;

.'. the polar of A passes through P : Ex. 2, p. 392.

that is, the point P lies on the polar of A.

II. To shew that any point on the polar of A satisfies the given
conditions.

Let BC be the polar of A, and let P be any point on it.

Draw tangents PH, PK, and let HK be the chord of contact.

Now from Ex. 1, p. 391, we know that the chord of contact HK
is the polar of P,

and we also know that the polar of P must pass through A ; for P is
on BC, the polar of A : Ex. 2, p. 392.

that is, H K passes through A.

.'. P is the point of intersection of tangents drawn at the ex-
tremities of a chord passing through A.

From I. and II. we conclude that the required locus is the polar
of A.

Note. If A is without the circle, the theorem demonstrated in
Part I. of the above proof still holds good ; but the converse theorem
in Part II. is not true for all points in BC. For if A is without the
circle, the polar BC will intersect it ; and no point on that part of
the polar which is within the circle can be the point of intersection
of tangents.

We now see that

(i) The Polar of an external point with respect to a circle is the
chord of contact of tangents draionfrom it.

(ii) The Polar oj an internal point is the locus of the intersections
of tangents drawn at the extremities of all chcn'ds which pass through
it.

(iii) The Polar oj a point on the circumference is the tangent at
thai point.

394

EUCUDS ELEMENTS.

The following theorem is known as the Harmonic Property of
Pole and Polar.

4. Any straight line drawn thrcnigh a point is cut JiarmonicaUy
by the pointy its polai'y and the circumference of the circle.

Let AHB be a circle, P the given
point and H K its polar ; let Paqh be any
straight line drawn through P meeting
the polar at q and the 0*=^ of the circle
at a and b.

Then shall P, a, q, b 6e a harmonic
range.

In the case here considered, P is an
external point.

Join P to the centre O, and let PO
cut the O*^" at A and B : let the polar of
P cut the O™ at H and K, and PO at Q.

Join Qa, Q6, Oa, OH, 06, PH.
Then PH is a tangent to the O AHB.
From the similar triangles OPH, HPQ,

OP:PH = PH:PQ.
.-. PQ. P0=PH2

= Pa . Ph.
:. the points O, Q, a, h are concyclic :

.-. the L aQA = the L abO Ex. 5, p. 241.

= the Z- Oah i. 5.

= the Z- 0Q&, in the same segment.

And since QH is perp. to AB,
.-. the L aQH = the L 6QH.
.*. Qq and QP are the internal and external bisectors of the L aQJb :

/. P, a, g, 6 is a harmonic range. Ex. 1, p. 385.

The student should investigate for himself the case when P is an
internal point.

Conversely, it may he sheimi that if through a fixed point P any
secant is drawn cuiting the circumference of a given circle at a and b,
and ifqis the harmonic conjugate of P with respect <o a, b ; then the
locus ofqis the polar of P with respect to the given circle.

Ex. 1, p. 391.

DEFINITION.

A triangle so related to a circle that each side is the
polar of the opposite vertex is said to be self-cox^ugate
with respect to the circle.

POLE AND POLAR. 395

EXAMPLES ON POLE AND POLAR.

1. The straight line which joins any two points is the polar with
respect to a given circle of the point of intersection of their polars,

2. The point of intersection of any two straight lines is the pole of
the straight line which joins their poles.

3. Find the locus of the poles of all straight lines which pass
through a given point.

4. Find the locus of the poles^ with respect to a given circle, of
tangents drawn to a concentric circle.

5. If ttoo circles cut one another orthogonally and PQ he any
diameter of one of them ; shew that the polar of P with regard to the
other circle passes through Q.

6. If tvx) circles cut one another orthogonally ^ the centre of each
circle is the pole of their common chord with respect to the other circle.

7. Any two points subtend at the centre of a circle an angle equxd
to one of the angles formed by the polars of the given points.

8. O is the centre of a given circle, and AB a fixed straight line.
P is any poi^it in AB ; Jind the locu>s of the point inverse to P vnth

respect to the circle.

9. Given a circle^ and a fixed point O on its circumference : P is
any point on the circle . find the locus of the point inverse to P vnth
respect to any circle whcse centre is O.

10. Given two points A and B, and a circle whose centre is O ;
shew that the rectangle contained by OA and the perpendiciUar from B
on the polar of A is equal to the rectangle contained by OB and the
perpendicular from A on the polar of B.

11. Four points A, B, C, D are taken in order on the circun^erence
of a circle; DA, CB intersect at P, AC, BD at Q, and BA, CD m R ;
shew that the triangle PQR is self-conjugate with respect to the circle,

12. Give a linear construction for finding the polar of a given
point with respect to a given circle. Hence find a linear construction
for drawing a tangent to a circle from an external point.

13. If a triangle is self conjugate with respect to a circle, the
centre of the circle is at the orthocentre of the triangle.

14. The polars, with respect to a given circle, of the four points of
a harmonic range form a harmonic pencil : and conversely.

396

Euclid's elements.

IV. ON THE RADICAL AXIS.

1. To find the locus of points from which the tangents dravm to
iwo given circles are equal.

Let A and B be the centres of the given circles, whose radii are a
and h ; and let P be any point such that the tangent PQ drawn to
the circle (A) is equal to the tangent PR drawn to the circle (B).

It is required to find the locus of P.

Join PA, PB, AQ, BR, AB ; and from P draw PS perp. to AB.

Then because PQ= PR, .'. PQ2=PR2.
But PQ2= PA2 - AQ2 ; and PR2= PB^ - BR2 ; i. 47.

.-. PA2-AQ2=PB2-BR2;
that is, PS2 + AS2 - «« = ps2 + SB^ - 62 ; i. 47.

or, AS2-a2=SB2-62.

Hence AB is divided at S, so that AS^ - SB2=a2 _ 52 .

/. S is di, fixed point.

Hence all points from which equal tangents can be drawn to the
two circles lie on the straight line which cuts AB at rt. angles, so
that the difference of the squares on the segments of AB is equal to
the diiFerence of the squares on the radii.

Again, by simply retracing these steps, it may be shewn that in
Fig. 1 every point in SP, and in Fig. 2 every point in SP exterior to
the circles, is such that tangents drawn from it to the two circles are
equal.

Hence we conclude that in Fig. 1 the whole line SP is the required
locus, and in Fig. 2 that part of SP which is without the circles.

In either case SP is said to be the Radical Axis of the two circles.

THE RADICAL AXIS. 397

Corollary. If the. circles cut one another as in Fig. 2, it is clear
that the Radical Axis is identical ivith the straight line which passes
through the points of intersection of the circles ; for it follows readily
from IIL 36 that tangents drawn to two intersecting circles from any
point in the common chord produced are equal.

2. The Radical Axes of three circles taken in pairs are concurrent.

Let there be three circles whose centres are A, B, C.
Let OZ be the radical axis of the 0" (A) and (B) ;
and OY the Radical Axis of the 0" (A) and (C), O being the point
of their intersection.

Then shall the radical axis of the 0* (B) and (C) pa^s through O.

It will be found that the point O is either without or within all
the circles,

L When O is without the circles.

Then because O is a point on the radical axis of (A) and (B) ; Hyp,

.'. OP=OQ.
And because O is a point on the radical axis of (A) and (C), Hyp.

:. OP=OR;

/. OQ=OR;

.*. O is a point on the radical axis of (B) and (C) ;

that is, the radical axis of (B) and (C) passes through O.

II. If the circles intersect in such a way that O is within them
all ;

the radical axes are then the common chords of the three circles
taken two and two ; and it is required to prove that these commoa
(Chords are concurrent. This may be shewn indirectly by in. 35.

Definition. The point of intersection of the radical axes
of three circles taken in pairs is called the radical centre.

398

Euclid's elements.

3. To draw the radical axis of two given circles.

Let A and B be the centres of the given circles.

It is required to draw their radical axis.

If the given circles intersect, then the st. line drawn through their
points of intersection will be the radical axis. [Ex. 1, Cor. p. 397.]
But if the given circles do not intersect,

describe any circle so as to cut them in E, F and G, H.

Join EF and HG, and produce them to meet in P.

Join AB ; and from r draw PS perp. to AB.

Then PS shall he the radical axis of the 0" (A), (B).
[The proof follows from iii. 36 and Ex. 1, p. 396.]

Definition. If each pair of circles in a given system
have the same radical axis, the circles are said to be co-axaL

EXAMPLES ON THE RADICAL AXIS.

1. Shew thai the radical axis of two circles bisects any one of their
common tangents,

2. If tangents are drawn to two circles from any point on their
radical axis ; shew that a circle described with this point as centre and
any one of the tangents as radius^ cuts both the given circles ortho-
gonally,

3. O is the radical centre of three circles^ and from O a tangent
OT is drawn to any one of them : sJieio that a circle whose centre is O
and radiiis OT exits all the given circles orthogonally.

4. If three circles touch one another^ taken two and two, shew that
their common tangents at the points of contact are concurrent.

EXAMPLES ON THE RADICAL AXIS. 399

6. If circles are descnhed on the three sides of a triangle as
diameter, their radical centre is the orthocentre of the triangle.

6. All circles which pass through a fixed point and cut a given
circle orthogojiaJly, pass through a second fixed point,

7. Find the locus of the centres of all circles which pass through a
given point and cut a given circle orthogonally.

8. Describe a circle to pass through two given points and cut
a given circle orthogonally.

9. Find the locus of the centres of all circles which cut two given
circles orthogonally.

10. Describe a circle to pass through a given point and cut two
given circles orthogonally.

11. The difference q/ the squares on the tangents drawn from any
point to tioo circles is equal to twice the rectangle contained by the
straigJU line joining their centres and the perpendicular from the given
point cm their radical axis.

12. In a system of co-axal circles which do not intersect, any point
is taken on the radical a>xis ; shew that a circle described from this
poird as centre, with radius equal to the tangent drawn from it to any
one of the circles, will meet the line of centres in two fixed points.

[These fixed points are called the Limiting Points of the system.']

13. /» a system of co-axal circles the two limiting points and the
points in tvhich any one circle of the system cuts the line of centres
form a hai'monic range.

14. In a system of co-axal circles a limiting 2wint has the same
polar with regard to all the circles of the system.

15. If two circles are orthogonal any diameter of one is cut
harmonically by the other.

V. ON TRANSVERSALS.

In the two following theorems we are to suppose that
the segments of straight lines are expressed numerically
in terms of some common unit ; and the ratio of one such
segment to another will be denoted by the fraction of which
the first is the numerator and the second the denominator.

H.S.E. 2o

400 EUCLID'S ELEMENTS.

Definition. A straight line drawn to cut a given
system of lines is called a transversal.

1. If three concurrent straight lines are drawn froni the angular
points of a triangle to meet the opposite sides, then the product of three
alternate segments taken in order is equal to the product of the other
three segments.

Let AD, BE, CF be drawn from the vertices of the A ABC to
intersect at O, and cut the opposite sides at D, E, F.

Then sJiall BD . CE . AF = DC . EA . FB.

Now the A" AOB, ACC have a common base AC; and it may be
shewn that

BD : DC = the alt. of A AOB : the alt. of A AOC ;

BD A AOB

• UC~AAOC'

. ., , CE A BOC

similarly.

and

EA~A BOA'

AF _ A COA
FB~A COB'

Multiplying these ratios, we have

BD CE AF

or.

DC EA FB~^'
BD . CE . AF = DC . EA . FB. Q.E.a

Note. The converse of this theorem, which may be proved in-
directly, is very important : it may be enunciated thus :

If three straight lines drawn from the vertices of a triangle cut the
opposite sides so that the product of three alternate segments taken in
order is equal to the product of the other three, then the three straight
lines are concurrent.

That is, if BD . CE . AF = DC . EA . FB,

then AD, BE, CF are concurrent.

ON TRANSVERSALS. 401

2. If a transversal is dratmi to cut the sides, or the sides produced,
of a trianghy the product of three cUtemate segments taken in order is
equal to the product of the other three segments.

B CD^

Let ABC be a triangle, and let a transversal meet the sides BC,
CA, AB, or these sides produced, at D, E, F.

Then shall BD . CE . AF = DC . EA . FB.

Draw AH par^ to BC, meeting the transversal at H.
Then from the similar A» DFB, HAF,

FB"

"AF*

and from the similar H

.« DCE, HAE,

DC"

"HA*

.*.,byi

multiplication.

BD
FB*

CE
DC"

EA
AF^

that is,

BD
DC

.CE.
. EA.

.AF
.FB"

= 1,

or,

.CE

.AF =

= DC.

EA.

FB.

Q.E.D.

Note. In this theorem the transversal must either meet two
sides and the third side produced, as in Fig. 1 ; or all three sides
produced, as in Fig. 2.

The converse of this theorem may be proved indirectly :

If three points are taken in tivo sides of a triangle and the third
side producedy or in all three sides produced, so that the product oj
three alteimate segments taken in order is equal to the product of the
other three segments, the three points are collinear.

402 Euclid's elements.

definitions.

1. If two triangles are such that three straight lines
joining corresponding vertices are concurrent, they are said
to be co-polar.

2. If two triangles are such that the points of inter-
section of corresponding sides are coUinear, they are said
to be co-axial.

The propositions given on pages 111-114 relating to the con-
currence of straight lines in a triangle, may be proved by the method
of transversals, and in addition to these the following important
theorems may be established.

Theorems to be proved by Transversals.

1. TVie straight lines which join the vertices of a triangle to the
points of contact of the inscribed circle {or any of the three escribed
circles) are concurrent.

2. The middle points of the diagonals of a complete qvudrilateral
are collinear. [See Def. 4, p. 387.]

3. Go-polar triangles are also co-axial ; and conversely co-axial
triangles are also co-polar.

4. The six centres of similitiide of three circles lie three by three
on four straight lines.

MISCELLANEOUS EXAMPLES ON BOOK VI. 403

MISCELLANEOUS EXAMPLES ON BOOK VI.

1. Through D, any point in the base of a triangle ABC,
straight lines UE, DF are drawn parallel to the sides AB, AC, and
meeting the sides at E, F : shew that the triangle AEF is a mean
proportional between the triangles FBD, EDC.

2. If two triangles have one angle of the one equal to one
angle of the other, and a second angle of the one supplementary to
a second angle of the other, then the sides about the third angles
are proportional.

3. AE bisects the vertical angle of the triangle ABC and meets
the base in E ; shew that if circles are described about the triangles
ABE, ACE, the diameters of these circles are to each other in the
same ratio as the segments of the base.

4. Through a fixed point O draw a straight line so that the
parts intercepted between O and the perpendiculars drawn to the
straight line from two other fixed points may have a given ratio.

5. The angle A of a triangle ABC is bisected by AD meeting BC
in D, and AX is the median bisecting BC : shew that XD has the
same ratio to XB as the difference of the sides has to their sum.

6. AD and AE bisect the vertical angle of a triangle internally
and externally, meeting the base in D and E ; shew that if O is the
middle point of BC, then OB is a mean proportional between OD
and OE.

7. P and Q are fixed points ; AB and CD are fixed parallel
straight lines ; any straight line is drawn from P to meet AB at M,
and a straight line is drawn from Q parallel to PM meeting CD at
N : shew that the ratio of PM to QN is constiint, and thence shew
that the straight line through M and N passes through a fixed point.

8. If C is the middle point of an arc of a circle whose chord is
AB, and D is any point in the conjugate arc ; shew that

AD + DB : DC :: AB : AC.

9. In the triangle ABC the side AC is double of BC. If CD,
CE bisect the angle ACB internally and externally meeting AB in D
and E, shew that the areas of the triangles CBD, ACD, ABC, CDE
are as 1, 2, 3, 4.

10. AB, AC are two chords of a circle ; a line parallel to the
tangent at A cuts AB, AC in D and E respectively : shew that the
rectangle AB, AD is equal to the rectangle AC, AE.

404 EUCLID'S ELEMENTS.

11. If from any point on the hypotenuse of a right-angled
triangle perpendiculars are drawn to the two sides, the rectangle
contained by the segments of the hypotenuse will be equal to the
sum of the rectangles contained by the segments of the sides.

12. D is a point in the side AC of the triangle ABC, and E'is a
point in AB. If BD, CE divide each other into parts in the ratio
4 : 1, then D, E divide CA, BA in the ratio 3 : 1.

13. If the perpendiculars from two fixed points on a straight
line passing between them be in a given ratio, the straight line
must pass through a third fixed point.

14. PA, PB are two tangents to a circle; PCD any chord
through P : shew that the rectangle contained by one pair of opposite
sides of the quadrilateral ACBD is equal to the rectangle contained
by the other pair.

15. A, B, C are any three points on a circle, and the tangent at
A meets BC produced in D : shew that the diameters of the circles
circumscribed about ABD, ACD are as AD to CD.

16. AB, CD are two diameters of the circle ADBC at right
angles to each other, and EF is any chord ; CE, CF are drawn
meeting AB produced in G and H ; prove that

the rect. CE, HG = the rect. EF, CH.

17. From the vertex A of any triangle ABC draw a line meeting
BC produced in D so that AD may be a mean proportional between
the segments of the base.

18. Two circles touch internally at O ; AB a chord of the larger
circle touches the smaller in C which is cut by the lines OA, OB in
the points P, Q : shew that OP : OQ : : AC : CB.

19. AB is any chord of a circle ; AC, BC are drawn to any
point C in the circumference and meet the diameter perpendicular
to AB at D, E : if O is the centre, shew that the rect. OD, OE is
equal to the square on the radius.

20. YD is a tangent to a circle drawn from a point Y in the
diameter AB produced ; from D a perpendicular DX is drawn to the
diameter ; shew that the points X, Y divide AB internally and ex-
ternally in the same ratio.

21. Determine a point in the circumference of a circle, from
which lines drawn to two other given points shall have a given
ratio.

MISCELLANEOUS EXAMPLES ON BOOK VL 405

22. O is the centre and OA a radius of a given circle, and V
is a fixed point in OA ; P and Q are two points on the circum-
ference on opposite sides of A and equidistant from it ; QV is
produced to meet the circle in L ; shew that, whatever be the length
of the arc PQ, the chord LP will always meet OA produced in a
fixed point.

23. EA, EA' are diameters of two circles touching each other
externally at E ; a chord AB of the former circle, when produced,
touches tne latter at C\ while a chord A'B' of the latter touches the
former at : prove that the rectangle, contained by AB and A'B', is
four times as great as that contained by BO" and B'C.

24. If a circle be described touching externally two given
circles, the straight line passing through the points of contact will
intersect the line of centres of the given circles at a fixed point.

25. Two circles touch externally in ; if any point D be taken
without them so that the radii AC, BO subtend equal angles at D,
and DE, DF be tangents to the circles, shew that DC is a mean
proportional between DE and DF.

26. If through the middle point of the base of a triangle any
line be drawn intersecting one side of the triangle, the other pro-
duced, and the line drawn parallel to the base from the vertex, it
will be divided harmonically.

27. If from either base angle of a triangle a line be drawn
intersecting the median from the vertex, the opposite side, and the
line drawn parallel to the base from the vertex, it will be divided
harmonically.

28. Any straight line drawn to cut the arms of an angle and its
internal and external bisectors is cut harmonically.

29. P, Q are harmonic conjugates of A and B, and C is an
external point ; if the angle PCQ is a right angle, shew that CP, CQ
are the internal and external bisectors of the angle ACB.

30. From C, one of the base angles of a triangle, draw a straight
line meeting AB in G, and a straight line through A parallel to the
base in E, so that CE may be to EG in a given ratio.

31. P is a given point outside the angle formed by two given
lines AB, AC ; shew how to draw a straight line from P such that
the parts of it intercepted between P and the lines AB, AC may
have a given ratio.

406 EUCLID*S ELEMENTS.

32. Through a given point within a given circle, draw a straight
line such that the parts of it intercepted between that point and the
circumference may have a given ratio. How many solutions does
the problem admit of ?

33. If a common tangent be drawn to any number of circles
which touch each other internally, and from any point of this
tangent as a centre a circle be described, cutting the other circles ;
and if from this centre lines be drawn through the intersections
of the circles, the segments of the lines within each circle shall be
equal.

34. APB is a quadrant of a circle, SPT a line touching it at P ;
C is the centre, and PM is perpendicular to CA ; prove that

the A SCT : the A ACB : : the A ACB : the A CMP,

35. ABC is a triangle inscribed in a circle, AD, AE are lines
drawn to the base BC parallel to the tangents at B, C respectively ;
shew that AD=AE, and BD : CE : : AB^ : AC^.

36. AB is the diameter of a circle, E the middle point of the
radius OB ; on AE, EB as diameters circles are described ; PQL is
a common tangent touching the circles at P and Q, and AB produced
at L: shew that BL is equal to the radius of the smaller circle.

37. The vertical angle C of a triangle is bisected by a straight
line which meets the base at D, and is produced to a point E, such
that the rectangle contained by CD and uE is equal to the rectangle
contained by AC and CB : shew that if the base and vertical angle
be given, the position of E is invariable.

38. ABC is an isosceles triangle having the base angles at B
and C each double of the vertical angle : if BE and CD bisect the
base angles and meet the opposite sides in E and D, shew that DE
divides the triangle into figures whose ratio is equal to that of AB
to BC.

39. If AB, the diameter of a semicircle, be bisected in C, and on
AC and CB circles be described, and in the space between the three
circumferences a circle be inscribed, shew that its diameter will be
to that of the equal circles in the ratio of 2 to 3.

40. O is the centre of a circle inscribed in a quadrilateral ABCD ;
a line EOF is drawn and making equal angles with AD and BC, and
meeting them in E and F respectively : shew that the triangles
AEO, BOF are similar, and that

AE : ED = CF : FB.

MISCELLANEOUS EXAMPLES ON BOOK VI. 407

41. From the last exercise deduce the following : The inscribed
circle of a triangle ABC touches AB in F ; XOY is drawn through
the centre making equal angles with AB and AC, and meeting them
in X and Y respectively : shew that BX : XF = AY : YC.

42. Inscribe a square in a given semicircle.

43. Inscribe a square in a given segment of a circle.

44. Describe an equilateral triangle equal to a given isosceles
triangle.

45. Describe a square having given the difference between a
diagonal and a side.

46. Given the vertical angle, the ratio of the sides containing it,
and the diameter of the circumscribing circle, construct the triangle.

47. Given the vertical angle, the line bisecting the base, and
the angle the bisector makes with the base, construct the triangle.

48. In a given circle inscribe a triangle so that two sides may
pass through two given points and the third side be parallel to a
given straight line.

49. In a given circle inscribe a triangle so that the sides may
pass through the three given points.

60. A, B, X, Y are four points in a straight line, and O is such a
point in it that the rectangle OA, OY is equal to the rectangle OB, OX ;
if a circle is described with centre O and radius equal to a mean
proportional between OA and OY, shew that at every point on this
circle AB and XY will subtend equal angles.

51. O is a fixed point, and OP is any line drawn to meet a fixed
straight line in P ; if on OP a point Q is taken so that OQ to OP is
a constant ratio, find the locus of Q.

52. O is a fixed point, and OP is any line drawn to meet the
circumference of a fixed circle in P ; if on OP a point Q is taken so
that OQ to OP is a constant ratio, find tlie locus of Q.

53. If from a given point two straight lines are drawn including
a given angle, and having a fixed ratio, find the locus of the extremity
of one of them when the extremity of the other lies on a fixed
straight line.

54. On a straight line PAB, two points A and B are marked and
the line PAB is made to revolve round the fixed extremity P. C is
a fixed point in the plane in which PAB revolves ; prove that if CA
and CB be joined and the parallelogram CADB be completed, the
locus of D will be a circle.

408 Euclid's elements.

55. Find the locus of a point whose distances from two fixed
points are in a given ratio.

56. Find the locus of a point from which two given circles
subtend the same angle.

57. Find the locus of a point such that its distances from two
intersecting straight lines are in a given ratio.

58. In the figure on page 389, shew that QT, PT' meet on the
radical axis of the two circles.

59. ABC is any triangle, and on its sides equilateral triansles
are described externally : if X, Y, Z are the centres of their inscribed
circles, shew that the triangle XYZ is equilateral.

60. If S, I are the centres, and R, r the radii of the circum-
scribed and inscribed circles of a triangle, and if N is the centre of
its nine-points circle,

prove that (i) SP = R2 - 2Rr,

(ii) NI=iR-r.

Establish corresponding properties for the escribed circles, and hence
prove that the nine-points circle touches the inscribed and escribed
circles of a triangle.

SOLID GEOMETRY

EUCLID. BOOK XI

Definitions.

From the Definitions of Book I. it will bo remembered
that

(i) A line is that which has length, without breadth
or thickness.

(ii) A surface is that which has length and breadth,
without thickness.

To these definitions we have now to add :

(iii) Space is that which has length, breadth, and
thichiess.

Thus a line is said to be of one dimeiiBion ;

a surface is said to be of two dimen^ons ;
and space is said to be of three dimensions.

The Propositions of Euclid's Eleventh Book here given
establish the first principles of the geometry of space, or solid
gemnetry. They deal with the properties of straight lines
which are not all in the same plane, the relations which
straight lines bear to planes which do not contain those
lines, and the relations which two or more planes bear to
one another. Unless the contrary is stated the straight
lines are supposed to be of indefinite length, and the planes
of infinite extent.

Solid geometry then proceeds to discuss the properties
of solid figures, of surfaces which are not planes, and of
lines which cannot be drawn on a plane surface.

410 euclid's elbmisnts.

Lines and Planes.

1. A etrdight line is perpendicular to a plane when
it is perpendicular to every straight tine which meets it
in that plana

Note. It will be prmed n Propoa t on 4 that if a. straight lino
IB perpendicular to Ivjo straight lines which meet it in a plane, it is
also perpendicnlar.to evtry straight lino which meets it in that plane.

A atrught line drawn perpendicular to a plane is said to be a
normal to that plane.

2. The foot of the perpendicular let fall from a given
point on a plane is called the projectioii of tliat point on

3. The projection of a line on a plane is the locus of

the feet of perpendiculars drawn from all points in the
given line to the plane.

Thus in the above figure the line nh is the projection of the line
e on the plane PQ.

NoTB. It will be proved hereafter (see page 4461 that the pro-
iction of a straight line on a plane is also a straight line.

DEFINITIONS.

411

4. The inclination of a straight line to a plane is the
acute angle contained by that line and another drawn from
the point at which the first line meets the plane to the
point at which a perpendicular to the plane let fall from
any point of the first line meets the plane.

Thus in the above figure, if from any point X in the given
straight line AB, which intersects the plane PQ at A, a perpen-
dicular Xx is let fall on the plane, and the straight line Axb is drawn
from A through a:, then the inclination of the straight line AB to the
plane PQ is measured by the acute angle BA6. In other words : —

The inclination of a straight line to a plane is the acute angle
contained by the given straight line and its projection on the plane.

Axiom. If two surfaces intersect one another, they meet
in a line or lines.

5. The common section of two intersecting surfaces is
the line (or lines) in which they meet.

Note. It is proved in Proposition 3 that the common section of
two planes is a straight line.

Thus AB, the common section of the two planes PQ, XY is proved
to be a straight line.

EUCLID'S ELEMEKT&

6. One plane is perpendicular to another plane when
any straight line drawn in one of the planes perpendicular
to the common section is also perpendicular to the other
plane.

Thus in the above figare, the plane EB is perpendicular to
the plane CD, if any straight line PQ, drawn in the plane EB at
tight angles to the common section AB, is also at right angles to the
plane CD.

7. The inclination of a plane to a plane is the acute

angle contained by two straight lines drawn irom any point
in the common section at right angles to it, one in one
plane and one in the other.

Thus in the adjoining figure,
the straight line AB is the com-
mon section of the two inter-
eecting planes BC, AD : and
from Q, any point in AB, two
straight lines QP, QR are drawn
perpendicular to AB, one in each
plane : then the inclination of
the two planes ia measured by
the acute angle PQR.

Note. This definition assumes that 'the angle PQ,R is of constant
magnitude whatei'er point Q is taken in AB : the truth of which
assumption is proved in Proposition 10.

The angle formed by the intersection of two planes is called a
dUiedraloi^e.

It may be proved that two planes are perpendicular to one another
when the dihedral angle formed by them is a right angle.

DEFINITIONS. 41S

Parallel planes are such as do not meet when pro-

9. A straight line is parallel to a plane if it does not
meet the plane when produced.

10. The angle between two straight hnea which do not
meet is the angle contained by two intersecting straight Hues
respectively parallel to the two non -intersecting lines.

Thus if AB and CD are two
straight lines which do nob meet,
and ab, be are two intersecting Unes
parallel respectively to AB and CD ;
then the angle between AB and CD
is measured by the angle aic.

11. A solid angle is that which is ma^le by three or
more plane angles which have a common \ertex, hut are
not in the same plane.

A solid angle made by three
plane angles is said to be tiltiedral ;
if made by more than three, it is
said to be pDlyhsdcsl.

A solid angle is sometimes called

12. A solid figure is any portion of space bounded by
one or more surfaces, plane or curved

These surfaces are called the Tacas of the solid and the inter
sections of adjacent faces are called ei^es

iU EUCLID'S ELEMENTS.

POLYHEDRA.

13. A polyhedron is a solid figure bounded by plane
faces.

Note. A plane rectilineal figure must at leaat have three aidea ;
oifour, if two of the sides are parallel. A polyhedron must at least
have/oui' faces ; or, if two faces are parallel, it must at least have
_fiae faces.

14 A pnsm is a solid figure bounded by plane faces,
of which two that ire oppo ite are similar and equal
polygons n p'srallel planes a d tho other faces are parpl-
lelograma

Ijl

The polygons are called the enda of the pnsm. A prit

obliqoe.

15. A parallelepiped is a solid figure bounded by three
pairs of parallel plane faces.
Fig 1

Fig 2.

lili

n fig. 1 , or oUiqwe as in
to a rect[ina;nlar paral-
lelepiped IV hose length breadth, and tliicknesa are not all equaL

DEFINITIONS.

415

16. A pynunid ia a solid figure bounded by plane
£ic6s, of which one is a polygon, and thq rest are triangles
having as bases the aides of the polygon, and as a common
vertex some point not in the plane of the polygon.

The polygon ia calls i the baaa of the pyramid

A pyramid having for ita base a regidar polygon is said to be
rlglit when the vertex lies in the etraignt line drawn perpendicular
to the baae from ita central pomt (the centre of its inscribed or
circumacribed circle)

17. A tetrahedion is a pyramid
on a triangular base : it is thus con-
tained by fmir triangular faces.

18. Polyhedra are classified according to the number
of their /aces :

thoH a hexalMdron has six facea ;
an oetobedroD has eight faces ;
a dodecahednin haa twdve faces.

19. Similar polyhedia are such as have all their solid
angles equal, each to each, and are bounded by the same
number of similar faces.

20. A polyhedron is regular when ite faces are similar
and equal regulai

EtTCUD's ELEMENTS.

21. It will be proved (see page 451) that there can
only be^iw regular polyhedra.
They are defined as follows : —

(i) A regnlar tetraliedroii is
a solid figure bounded h\ Jmir
plane faces, which are equal and
equilateral triangles.

(ii) A cube is a solid figure
bounded by six plane faces, which
are equal squares. .

(iii) A regular octahedron is a
solid figure bounded by eiqht plane
faces, wnich are equal and equilateral
triangles.

(n ) A regular dodecahedron is .
a solid figure bounded bj' luvlve plane
faces, which are equal and regular

S?) A regular icosalkedron is
figure bounded by hbenty
plane faces, which are equal and
equilateral triangles.

Solids of Revolution.

22o. A sphere is a solid figure described by the revo-
lution of a semicircle about its diameter, which remains
fixed.

I straight line about which the
me as the centre of the semi-

inalogous to that given tor

22i. A sphere is a solid figure contained by one surface,
which is such that all straight lines drawn from a certain
point within it to the surface are equal : thia point is called
the centre of the sphere.

It will be seen that the surface of a sphere is the locus of a point
which moves in fpace so that its dlstaiice from a certain fixed point
(the oentw) i

EUCUO'S ELEMENTS.

33. A lUsht cylinder is a solid
figure described by the revolution of
a rectangle about one of its eidee
which remaim fixed.

The wds of the cyliuder is the fixed straight line about which
the rectangle revolves.

The Imses, or ends, of the cylinder are the circular faces described
b; the two revolving opposite sides of the rectangle.

24. A right cone ie a solid ligure
described by the revolution of a right-
angled triangle about one of the sides
containing the right angle which re-
mains fixed.

The rnzla of the cone is the fised straight line about which the
triangle revolves.

The base of the cone is the circular face described by that side

25. Similar cones and cylinders are those which have
their axes and the diameters of their bases proportionals.

BOOK XI. PROP. 1. 419

Proposition 1. Theorem.

One part of a straight line cannot be in a plans and another
part ovicide it.

If possible, let AB, part of the st. line ABC, be in the
plane PQ, and the part BC outside it.

Then since the st. line AB is in the plane PQ,

.'. it can be produced in that plane, i. Post, 2.

Produce AB to D ;
and let any other plane which passes through AD be turned
about AD until it passes also through C.

Then because the points B and C are in this plane,

.". the St. line BC is in it : i. Def, 7.

.'. ABC and ABD are in the same plane and are both
St. lines ; which is impossible. I. Def, 4.

.*. the st line ABC has not one part AB in the plane PQ,
and another part BC outside it. Q.E.D.

Note. This proposition scarcely needs proof, for the truth of it
follows immediately from the de6nitions of a straight line and a
plane.

It shonld be observed that the method of proof used in this and
the next proposition rests upon the following axiom :

If a plane of unlimited extent turns about a fixed straight line cw
an aaciSf it can he made to pass through any point in space.

420 EUCLID'S ELEMENTS.

Proposition 2. Theorem.

Any two intersecting straight lines are in one plane:
and any three straight lines, of which each pair intersect one
another, are in one plans.

Let the two st. lines AB and CD intersect at E ;
and let the st. line BC be drawn cutting AB and CD at B
and C.

Then (i) AB and CD shall lie in one plans.
(ii) AB, BC, CD shall lie in one plane,

(i) Let any plane pass through AB ;

and let this plane be turned about AB until it passes
through C.

Then, since C and E are points in this plane,
.*. the whole st. line CED is in it. I. Def. 7 and XI. L
That is, AB and CD lie in one plane.

(ii) And since B and C are points in the plane which
contains AB and CD,

.*. also the st. line BC lies in this plane. Q.E.D.

Corollary. One, and only one, plane can be made to pass
through two given intersecting straight lines.

Hence the position of a plane is fixed,

(i) if it passes through a given straight Hne and a given point
outside it ; Ax. -p. 419.

(ii) if it passes through two intersecting straight lines ; xi. 2.

(iii) if it passes through three points not collinear ; xi. 2.

(iv) if it passes through two parallel straight lines, i. Def. 35.

book. xi. prop. 3.
Proposition 3. Theorem.

421

If tvx) planes cvi one another^ their common section is a
straight line.

Let the two planes XA, CY cut one another, and let BD be
their common section.

ITien shall BD be a straight line.

For if not, from B to D in the plane XA draw the st. line
BED ;

and in the plane CY draw the st. line BFD.

Then the st. lines BED, BFD have the same extremities;

.*. they include a space ;
but this is impossible. Ax. 10.

,'. the common section BD cannot be otherwise than a st.
line.

Alternative Proof.

Let the planes XA, CY cut one another, and let B and D be
two points in their common section.

Then because B and D are two points in the plane XA,

.*. the St. line joining B, D lies in that plane. I. Def. 7.

And because B and D are two points in the plane CY,

.'. the St. line joining B, D lies in that plane.

Hence the st. line BD lies in both planes,

and is therefore their common section.

That is, the common section of the two planes is a straight

line. Q.E.D.

422

EUCLID S ELEMENTS.

Proposition 4. Theorem. [Alternative Proof.]

If a straight line is perpendicular to each of two straight lines
' at their point of intersection, it shall also be perpendicular to the
plane in which they lie.

Let the straight line AD be perp. to each of the st. lines
AB, AC at A their point of intersection.

Then shall AD he perp. to ths plane in which AB and AC lie.

Produce DA to F, making AF equal to DA.

Draw any st. line BC in the plane of AB, AC, to cut
AB, AC at B and C ;

and in the same plane draw through A any st. line AE to cut
BC at E.

It is required to prove that AD is perp. to AE. XI. Def. 1.
Join DB, DE, DC ; and FB, FE, FC.

Then in the A" BAD, BAF,

because DA = FA, Constr,

and the common side AB is perp. to DA, FA ;

.-. BD = BF. I. 4.

Similarly CD = CF.

Now if the A BFC be turned about its base BC until the
vertex F comes into the plane of the A BDC,

then F will coincide with D,
since the conterminous sides of the triangles are equal, i. 7

.-. EF will coincide with ED,
that is, EF=ED.

BOOK XI. PROP. 4.

423

Hence in the A' DAE, FAE,

since DA, AE, ED = FA, AE, EF respectively,

.*. the z_ DAE = the l FAE.

That is, DA is perp. to AE.

Similarly it may be shewn that DA is perp. to ewry
St. line which meets it in the plane of AB, AC ;

.'. DA is perp. to this plane. Q.E.D.

Proposition 4. Theorem. [Euclid's Proof.]

If a straight line is perpendicular to each of two straight lines
at their point of intersection, it shall also be perpendicvlar to the
pla/ne in which they lie.

Let the st. line EF be perp. to each of the st. lines
AB, DC at E their point of intersection.

Then shall EF be also perp. to the plane XY, in which AB and
DC lie.

Make EA, EC, EB, ED all equal, and join AD, BC.
Through E in the plane XY draw any st. line cutting
AD and BC in G and H.

Take any pt. F in EF; and join FA, FG, FD, FB, FH, FC.

Then in the A" AED, BEC,

because AE, ED = BE, EC respectively, Constr,

and the l AED = the l BEC ; i. 16.

/. AD = BC, and the i. DAE = the L CBE. L 4.

424

EUCLID'S ELEMENTS.

In the A» AEG, BEH,
because the l GAE = the l HBE,
and the l AEG = the l BEH,
and EA=EB;

.-. EG = EH, and AG = BH.

Again in the A" FEA, FEB,
because EA— EB,
and the common side FE is perp. to EA, EB ;
.-. FA=FB.
Similarly FC = FD,

Again in the A" DAF, CBF,
because DA, AF, FD = CB, BF, FC, respectively,
.-. the z. DAF = the z. CBF.

Proved.

I. 15.

Constr.

I. 26.

Hyp,
I. 4.

L 8.

And in the A" FAG, FBH,
because FA, AG = FB, BH, respectively,
and the z_ FAG = the l FBH, Proved.

.'. FG=FH. L 4.

Lastly in the A" FEG, FEH,
because FE, EG, GF = FE, EH, HF, respectively,

.-. the L. FEG = the l FEH ; I. 8.

that is, FE is perp. to GH.

Similarly it may be shewn that FE is perp. to every
St. line which meets it in the plane XY,

.'. FE is perp. to this plane. XI. Def. 1.

Q.E.D.

BOOK XI. PROP. 5.

425

Proposition 5. Theorem.

If a straight line is perpendicular to each of three concurrent
straight lines at their point of intersection, these three straight
lines shall be in one plane.

Let the straight line AB be perpendicular to each of
the straight lines BC, BD, BE, at B their point of inter-
section.

Then shall BC, BD, BE be in one plane.

Let XY be the plane which passes through BE, BD; XI. 2.
and, if possible, suppose that BC is not in this plane.

Let AF be the plane which passes through AB, BC ;
and let the common section of the two planes XY, AF be the
St. line BF. XI. 3.

Then since AB is perp. to BE and BD,
.*. AB is perp. to the plane containing BE, BD, namely the
plane XY ; XL 4.

and since BF is in this plane,
.-. AB is also perp. to BF. XI. Def 1.

But AB is perp. to BC ; Hyp-

. • . the L" ABF, ABC, which are in the same plane AF, are
both rt. angles ; which is impossible.

.*. BC is not outside the plane of BD, BE :
that is, BC, BD, BE are in one plane.

Q.E.D.

426

Euclid's elements.
Proposition 6. Theorem.

If two straight lines are perpendicular to the same plane, they
shall be parallel to one another.

Let the st. lines AB, CD be perp. to the plane XY.

Then shall AB and CD be pai\*

Let AB and CD meet the plane XY at B and D.

Join BD;
and in the plane XY draw DE perp. to BD, making DE
equal to AB.

Join BE, AE, AD.

Then since AB is perp. to the plane XY, Hyp.

.-. AB is also perp. to BD and BE, which meet it in that
plane ; xr. Def. 1.

that is, the l" ABD, ABE are rt. angles.
Similarly the /.' CDB, CDE are rt. angles.

Now in the A" ABD, EDB,
because AB, BD = ED, DB, respectively, Constr.
and the l ABD = the l EDB, being rt. angles ;

.-. AD = EB.

Again in the A' ABE, EDA,

because AB, BE = ED, DA, respectively,

and AE is common ;

.-. the L ABE = the L EDA.

I. 4.

I. 8.

♦Note. In order to shew that AB and CD are parallel, it is
necessary to prove that (i) they are in the same plane ^ (11) the angles
ABD, CDB, are supplementary.

BOOK XI. PROP. 7. 427

But the L ABE is a rt. angle ; Proved.

,'. the z- EDA is a rt. angle.
But the L EDB is a rt. angle by construction,
and the z. EDC is a rt. angle, since CD is perp. to the
plane XY. Hyp,

Hence ED is perp. to the three lines DA, DB, and DC ;
.*. DA, DB, DC are in one plane. xi. 5.

But AB is in the plane which contains DA, DB ; XI. 2.
.*. AB, BD, DC are in one plane.
And each of the l' ABD, CDB is a rt. angle ; Hyp,
.', AB and CD are par*. l. 28.

Q.E.D.

Proposition 7. Theorem.

If two straight lines are jparallel, the straight line which joins
any point in one to any point in the other is in the same plane as
the parallels,

A E B

Let AB and CD be two par* st. lines,

and let E, F be any two points, one in each st. line.

Uien shall the st, line which joins E, F be in the same plane
as AB, CD.

For since AB and CD are par*,
.'. they are in one plane. i. Def. 35.

And since the points E and F are in this plane,
.*. the St. line which joins them lies wholly in this plane.

I. Def, 7,
That is, EF is in the plane of the par*' AB, CD.

QE.D.

428

EUCLID'S ELEMENTS.

Proposition 8. Theorem.

If two straight lines are parallel^ and if one of them is per-
pendicular to a plane, then the other shall also be perpendicular
to the same plane.

Let AB, CD be two par* st. lines, of which AB is peq).
to the plane XY.

Then CD shall also be perp. to the same plane.

Let AB and CD meet the plane XY at the points B, D.

Join BD ;
and in the plane XY draw DE perp. to BD, making DE equal
to AB.

Join BE, AE, AD.

Then because AB is perp. to the plane XY, Syp-

.'. AB is also perp. to BD and BE, which meet it in that

plane; XL Def. 1.

that is, the l^ ABD, ABE are rt. angles.

Now in the A" ABD, EDB,
because AB, BD = ED, DB, respectively, Constr,
and the l ABD = the l EDB, being rt. angles ;

.-. AD = EB. L 4.

Again in the A" ABE, EDA,
because AB, BE = ED, DA, respectively,
and AE is common ;
.-. the L ABE = the l EDA. I. 8.

BOOK XI. PROP. 8. 429

But the L ABE is a rt. angle ; Proved.

.'. the L EDA is a rt. angle :
that is, ED is perp. to DA.

But ED is also perp. to DB : Constr,

.'. ED is perp. to the plane containing DB, DA. xi. 4.
And DC is in this plane ;
for both DB and DA are in the plane of the par'" AB, CD.

XI. 7.
.-. ED is also perp. to DC ; xi. Def. 1.

that is, the z_ CDE is a rt. angle.

Again since AB and CD are par\ Hyp.

and since the l ABD is a rt. angle,
.-. the L CDB is also a rt. angle. I. 29.

.'. CD is perp. both to DB and DE ;
.*. CD is also perp. to the plane XY, which contains
DB, DE. XI. 4.

Q.E.D.

EXERCISES.

1. The perpendicular is the least straight line that can be drawn
from an external point to a plane.

2. Equal straight lines drawn from an external point to a plane
are equally inclined to the perpendicular drawn from that point to
the plane.

3. Shew that two observations with a spirit-level are sufficient
to determine if a plane is horizontal : and prove that for this purpose
the two positions of the level must not be parallel.

4. What is the locus of points in space which are equidistant
from two fixed points ?

5. Shew how to determine in a given straight line the point
which is equidistant from two fixed points. When is this im-
possible ?

6. If a straight line is parallel to a plane, shew that any plane
passing through the given straight line will have with the given plane
a common section which is parallel to the given straight line.

430

EUCLID'S ELEMENTS.

Proposition 9. Theorem.

Two straight lines which are parallel to a third straight line
are parallel to one another.

\g a

/ D

Let the st. lines AB, CD be each par* to the st. line PQ.
Then shall AB he pan^ to CD.

Case I. If AB, CD and PQ are in one plane, the proposition
has already been proved. I. 30.

Case II. But if AB, CD and PQ are not in one plane,

in PQ take any point G ;
and from G, in the plane of the par'" AB, PQ, draw GH

perp. to PQ; I. 11.

also from G, in the plane of the par^ CD, PQ, draw GK

perp. to PQ. L 11.

Then because PQ is perp. to GH and GK, Constr.
.', PQ is perp. to the plane HGK, which contains them.

XL 4.

But AB is par* to PQ ; Hyp.

.'. AB is also perp. to the plane HGK. XL 8.

Similarly, CD is perp. to the plane HGK.

Hence AB and CD, being perp. to the same plane, are par*
to one another. xi. 6.

Q.E.D.

BOOK XI. PROP. 10.

431

Proposition 10. Theorem.

If two interceding straight lines are respectively parallel to
two other intersecting straight lines not in the same plane with
them, then the first pair and the second pair shall contain equal
angles.

Let the st. lines AB, BC be respectively par* to the st.
lines DE, EF, which are not in the same plane with them.

Then shall the l ABC = the l DEF.

In BA and ED, make BA equal to ED ;

and in BC and EF, make BC equal to EF.
Join AD, BE, CF, AC, DF.

Then because BA is equal and par* to ED,

Hyp, and Gonstr.
.-. AD is equal and par* to BE. L 33.

And because BC is equal and par* to EF,

.'. CF is equal and par* to BE. I. 33.

Hence AD and CF, being each equal and par* to BE, are equal

and par* to one another ; Ax, 1 and XI. 9.

hence it follows that AC is equal and par* to DF. I. 33.

Then in the A" ABC, DEF,
because AB, BC, AC = DE, EF, DF, respectively,

.-. the L ABC = the l DEF. I. 8.

Q.E.D.

n.S. B.

432

EUCLID S ELEMENTS.

Proposition U. Problem.

To draw a straight line perpendicular to a given plane from
a given point outside it

Let A be the given point outside the plane XY.
// is required to draw from A a st. line perp, to the plane XY.

Draw any st. line BC in the plane XY ;

and from A draw AD perp. to BC. I. 12.

Then if AD is also perp. to the plane XY, what was
required is done.

But if not, from D draw DE in the plane XY perp.

to BC

I. 11.
I. 12.

and from A draw AF perp. to DE.
Then AF shall be perp. to the plane XY.

Through F draw FH par^ to BC. I. 31.

Now because CD is perp. to DA and DE, Constr,
.'. CD is perp. to the plane containing DA, DE. XI. 4.

And HF is par* to CD;
HF is also perp. to the plane containing DA, DE. XI. 8.

And since FA meets HF in this plane,

.-. the L HFA is a rt. angle; XL Def 1.

that is, AF is perp. to FH.
And AF is also perp. to DE ; Constr.

/. AF is perp. to the plane containing FH, DE ;

that is, AF is perp. to the plane XY. Q.E.F.

BOOK XL PROP. 12.

433

Proposition 12. Problem.

To draw a straight line perpendicular to a given plane from
a given point in the plane.

D B

Let A be the given point in the plane XY.
It is required to draw from A a st. line perp. to the plane XY.

From any point B outside the plane XY draw BC perp.
to the plane. XL 11.

Then if BC passes through A, what was required is
done.

But if not, from A draw AD par^ to BC. L 31.

Then AD shall be the perpendicular required.

For since BC is perp. to the plane XY, Constr,

and since AD is par^ to BC, Constr,

.'. AD is also perp. to the plane XY. XL 8.

EXERCISES.

1. Equal straight lines drawn to meet a plane from a point
without it are equally inclined to the plane.

2. Find the locus of the foot of the perpendicular drawn from a
given point upon any plane which passes through a given straight
line.

3. From a given point A a perpendicular AF is drawn to a plane
XY; and from F, FD is drawn perpendicular to BC, any line in
that plane : shew that AD is also perpendicular to BC.

434

EUCLID S ELEMENTS.

Proposition 13. Theorem.

Only one perpendicular can be dravm to a given plane from
a given point either in the plane or outside it

\
\

/ ■

•

' /

r &

Case I. Let the given point A be in the given plane XY ;
and, if possible, let two perps. AB, AC be drawn from A to
the plane XY.

Let DF be the plane which contains AB and AC ; and
let the St. line DE be the common section of the planes DF
and XY. XL 3

Then the st lines AB, AC, AE are in one plane.

And because BA is perp. to the plane XY, Hyp,
.*. BA is also perp. to AE, which meets it in this plane \

XL Def. 1.
that is, the l BAE is a rt. angle.
Similarly, the l CAE is a rt. angle.
.'. the L^ BAE, CAE, which are in the same plane, are equal
to one another ; which is impossible.
.'. two perpendiculars cannot be drawn to the plane
XY from the point A in that plane.

Case II. Let the given point A be outside the plane XY.
Then two perp" cannot be drawn from A to the plane ;
for if there could be two, they would be par*, xi. 6.

which is absurd. Q.KD.

BOOK XI. P3M)P. U,

Proposition U. Theorem.

Planes io which the lamt straight line is
parallel to <me anoiher.

Xet tlie St. line A8 be perp. to each of the planes CO, EF.
Then shall the plomes CD, EF be pat'.
For if not, they will raeet when produced.
If possible, let the two planee meet, and let the at.
line GH be their common section. XI. 3.

In GH take any point K ;
and join AK, BK.
Then because AB is perp. to the plane EF,
.'. AB is also perp. to BK, which meets it in this plane ;
XI. Def. 1.
that is, the l. ABK is a rt. angle.
Similarly, the l BAK is a rt. angle,
.'.in the A KAB, the two z.*ABK, BAK are together equal to
two rt. angles ;

which is impossible. i. 17.

.■- the planes CD, EF, though produced, do not meet :
that is, they are par*. Q.E.D

436

EUCLID'S ELEMENTS.

Proposition 15. Theorem.

If two intersecting straight lines are parallel respectively to
two other intersecting straight lines which are not in the same
plane with them, then the plane containing the first pair shall be
parallel to the plane containing the second pair.

Let the st. lines AB, BC be respectively par^ to tbe
St. lines DE, EF, which are not in the same plane as
AB, BC.

nen shall the plane containing AB, BC he pan'' to the plane
containing DE, EF.

From B draw BG perp. to the plane of DE, EF; XL 11.
and let it meet that plane at G.
Through G draw GH, GK par^ respectively to DE, EF. I. 31.

Then because BG is perp. to the plane of DE, EF,
.*. BG is also perp. to GH and GK, which meet it in that
plane : XI. Def, 1.

that is, each of the l" BGH, BGK is a rt. angle.

Now by hypothesis BA is par^ to ED,
and by construction GH is par^ to ED;

.'. BA is par' to GH. XL 9.

And since the l BGH is a rt. angle ; Proved,
.', the L ABG is a rt. angle. L 29.

Similarly the z. CBG is a rt. angle.

BOOK XI. PROP. 16.

Then

m since BQ is perp. to each of the st. lines BA, BC,
.". BG is perp. to the plane containing them. XI. 4,
But BG is also perp. to the plane of ED, EF ; Ccmstr.
that is, BO is perp. to the two planes AC, DF ;

-■. these planes are par". xi. 14.

XI. 14.

Q.E.D.

If two mraUel planes are cut by a third plane, their
s&Avms with U shall be parallel.

Let the par* planes AB, CD be cut by the plane EFHQ,
and let the st. lines EF, GH be their common sections
with it.

ThenshaU EF, QH bepaf^.

For if not, EF and GH will meet if produced.

If possible, let them meet at K.

Then since the whole st. line EFK is in the plane AB, Xi, 1.

and K is a point in that line,

.'. the point K is in the plane AB.

Similarly the point K is in the plane CD.

Hence the planes AB, CD when produced meet at K ;

which is impossible, since they are par". Hyp.
.'. the st lines EF and QH do not meet ;
and they are in the same plane EFHG ;

.-. they are paH. i. Def. 36.

Q.K.D.

438 euclid's elements.

Proposition 17. Theorem.

Straight lines which are cut by parallel planes are cut pro-
portionaUy.

Let the st. lines AB, CD be cut by the three par* planes
GH, KL, MN at the points A, E, B, and C, F, D.

Then shall AE : EB : : CF : FD.

Join AC, BD, AD ;
and let AD meet the plane KL at the point X :

join EX, XF.

Then because the two par* planes KL, MN are cut by
the plane ABD,

.'. the common sections EX, BD are par\ XL 16.

And because the two par* planes GH, KL are cut by the
plane DAC,

.*. the common sections XF, AC are par*. XL 16.

Now since EX is par* to BD, a side of the A ABD,

.-. AE : EB :: AX : XD. VL 2.

Again because XF is par* to AC, a side of the A DAC,

.-. AX : XD :: CF : FD. VL 2.

Hence AE : EB :: CF : FD. V. 1.

Q.E.D.

Definition. One plane is perpendicular to another
plane, when any straight line drawn in one of the planes
perpendicular to their common section is also perpendicular
to the other plane. [Book XL Def, 6.]

book xi. prop. 18. 439

Proposition 18. Theorem.

If a straight Un^ is perpendicular to a plane, then aiery
plane which passes through the straight line is also perpendicuUtr
to the given plane.

Let the st. line AB be perp. to the plane XY ;
and let DE be any plane passing through AB.
Then shall the plane DE hepet-p. to the plave XV,
Let the at. line CE be the common section of the planes
XY, DE. XI. 3.

From F, any point in CE, draw FG in the plane OE
perp. to CE. i. 11

Then because AB is perp. to the plane XY, Hyp
.'. AB is also perp. to CE, which meets it in that plane,

XI. Def. 1,

that is, the z. ABF is a rt. angle.

But the - GFB is also a rt. angle ; Constr

.-. OF is par' to AB. i. 28.

And AB is perp. to the plane XY, Hyp.

.-. GF is a!so perp. to the plane XY. 3

Hence it has been shewn that any st. line QF drawn in

the plane DE perp. to the common section CE is also perp.

to the plane XY.

,-. the plane DE is perp. to the plane XY. XL Def. "
Q.K.D.

440

EUCLID'S ELEMENTS.

Proposition 19. Theorem.

If two intersecting planes are each perpendicular to a third
plane, their common section shall also he perpendicular to that
plane.

Let each of the planes AB, BC be perp. to the plane
ADC, and let BD be their common section.

Then shall BD he perp. to the plane ADC.

For if not, from D draw in the plane AB the st. line DE
perp. to AD, the common section of the planes ADB, ADC :

L 11.

and from D draw in the plane BC the st. line DF perp.
to DC, the common section of the planes BDC, ADC.

Then because the plane BA is perp. to the plane ADC,

Hyp.

and DE is drawn in the plane BA perp. to AD the common

section of these planes, Constr.

,'. DE is perp. to the plane ADC. XL Def. 6.

Similarly DF is perp. to the plane ADC.

.*. from the point D two st. lines are drawn perp. to the

plane ADC ; which is impossible. XI. 1 3.

Hence DB cannot be otherwise than perp. to the plane ADC.

Q.E.D.

BOOK XI. PROP. 20. 441

Proposition 20. Theorem.

Of the three plane angles which form a trihedral angle, any
two are together' greater than the third.

Let the trihedral angle at A be formed by the three
plane l* BAD, DAC, BAG.

Then shall any two of them, such as the iJ BAD, DAC, he together
greater than the third, the l BAG.

Case I. If the l BAG is less than, or equal to, either
of the L* BAD, DAG ;

it is evident that the <l' BAD, DAG are together greater than
the L BAG.

Case II. But if the l BAG is greater than either of the
z." BAD, DAG ;

then at the point A in the plane BAG make the l BAE equal
to the z. BAD ;

and cut off AE equal to AD.
Through E, and in the plane BAG, draw the st. line BEG
cutting AB, AG at B and G :

join DB, DG.

Then in the A" BAD, BAE,

since BA, AD = BA, AE, respectively, Constr.

and the l BAD = the l BAE ; Constr,
.-. BD = BE. I. 4.

Again in the A BDG, since BD, DG are together greater
than BG, I. 20.

and BD = BE, Proved,

.'. DG is greater than EG.

SUCLID S ELBUENTS.

And in the A- DAC, EAC,

because DA, AC = EA, AC respectively, Condr.

but DC is greater than EC ; Proved.

.■. the L DAC is greater than the l EAC. i. 25.

But the i. BAD = the l BAE ; Cottsir.

.-. the two i." BAD, DAC are together greater than the

L BAC. Q.E.D.

Proposition 21, Theorem.

£iwry (cmvex) sdid angle is formed by pUme angles v
■e loge^ier less than four right angles.

Let the solid angle at 8 be formed by the plane i.' ASB,
B8C, CSD, DSE, ESA.

Then shall the sum of these plane angles he leas than four
rt. angles.

BOOK XI. PROP. 21. 443

For let a plane XY intersect all the arms of the plane

angles on the same side of the vertex at the points A, B, C,

D, E : and let AB, BC, CD, DE, EA be the common sections

of the plane XY with the planes of the several angles.

Within the polygon ABODE take any point O ;

and join O to each of the vertices of the polygon.

Then since the l* SAE, SAB, EAB form the trihedral
angle A,
.*. the £.■ SAE, SAB are together greater than the l EAB ;

XI. 20.
that is,
the z-' SAE, SAB are together greater than the l* OAE, OAB.

Similarly,

the Z-' SBA, SBC are together greater than the l* OBA, OBC :
and so on, for each of the angular points of the polygon.

Thus bv addition,
the sum of the base angles of the triangles whose vertices
are at S, is greater than the sum of the base angles of
the triangles whose vertices are at O.

But these two systems of triangles are equal in number ;
.'. the sum of all the angles of the one system is equal to
the sum of all the angles of the other.

It follows that the sum of the vertical angles at 8 is less
than the sum of the vertical angles at O.

But the sum of the angles at O is four rt. angles ;
.*. the sum of the angles at 8 is less than four rt. angles.

Q.E.D.

Note. This proposition was not given in this form by Euclid,
who established its truth only in the case of trihedral angles. The
above demonstration, however, applies to all cases in which the
polygon ABODE is convex, but it must be observed that without
this condition the proposition is not necessarily true.

A solid angle is convex when it lies entirely on one side of each
of the infinite planes which pass through its plane angles. If this is
the case, the polygon ABODE will have no re-entrant angle. And it
is clear that it would not be possible to apply xi. 20 to a vertex at
which a re-entrant angle existed.

444 EUCLID'S ELEMENTS.

Exercises on Book XI.

1. Equal straight lines drawn to a plane from a point without
it have equal projections on that plane.

2. If S is the centre of the circle circumscribed about the triangle
ABC, and if SP is drawn perpendicular to the plane of the triangle,
shew that any point in SP is equidistant from the vertices of the
triangle.

3. Find the locus of points in space equidistant from three given
points.

4. From Example 2 deduce a practical method of drawing a
perpendicular from a given point to a plane, having given rufer,
compasses, and a straight rod longer than the required perpen-
dicular.

5. Give a geometrical construction for drawing a straight line
equally inclined to three straight lines which meet in a point, but
are not in the same plane.

6. In a gauche quadrilateral (that is, a quadrilateral whose sides
are not in the same plane) if the middle points of adjacent sides are
joined, the figure thus formed is a parallelogram.

7. AB and AC are two straight lines intersecting at right angles,
and from B a perpendicular BD is drawn to the plane in which they
are : shew that AD is perpendicular to AC.

8. If two intersecting planes are cut by two parallel planes, the
lines of section of the first pair with each of the second pair contain
equal angles.

9. If a straight line is parallel to a plane, shew that any plane
passing through the given straight line will intersect the given plane
in a line of section which is parallel to the given line.

10. Two intersecting planes pass one through each of two
parallel straight lines ; shew that the common section of the planes
is parallel to the given lines.

11. If a straight line is parallel to each of two intersecting
planes, it is also parallel to the common section of the planes.

12. Through a given point in space draw a straight line to
intersect each of two given straight lines which are not in the same
plane.

13. If AB, BC, CD are straight lines not all in one plane, shew
that a plane which passes through the middle point of each one of
them is parallel both to AC and BD.

14. From a given point A a perpendicular AB is drawn to a
plane XY ; and a second perpendicular AE is drawn to a straight
line CD in the plane XY : shew that EB is perpendicular to CD.

EXERCISES ON BOOK XI. 445

15. From a point A two perpendiculars AP, AQ are drawn one
to each of two intersecting planes : shew that the common section of
these planes is perpendicular to the plane of AP, AQ.

16. From A, a point in one of two given intersecting planes,
AP is drawn perpendicular to the first plane, and AQ perpendicular
to the second : if these perpendiculars meet the second plane at P
and Q, shew that PQ is perpendicular to the common section of the
two planes.

17. A, B, C, D are four points not in one plane, shew that the
four angles of the gauche quadrilateral ABCD [see Ex. 6, p. 444] are
together less than four right angles.

18. OA, OB, OC are three straight lines drawn from a given
point O not in the same plane, and OX is another straight line
within the solid angle formed by OA, OB, 00 : shew that

(i) the sum of the angles AOX, BOX, COX is greater than
half the sum of the angles AOB, BOG, COA.

(ii) the sum of the angles AOX, COX is less than the sum of
the angles AOB, COB.

(iii) the sum of the angles AOX, BOX, COX is less than the
sum of the angles AOB, BOC, COA.

19. OA, OB, OC are three straight lines forming a solid angle
at O, and OX bisects the plane angle AOB ; shew that the angle
XOC is less than half the sum of the angles AOC, BOC.

20. If a point is equidistant from the angles of a right-angled
triangle and not in the plane of the triangle, the line joining it with
the middle point of the hypotenuse is perpendicular to the plane of
the triangle.

21. The angle which a straight line makes with its projection on
a plane is less than that which it makes with any other straight line
which meets it in that plane.

22. Find a point in a given plane .such that the sum of its
distances from two given points (not in the plane but on the same
side of it) may be a minimum.

23. If two straight lines in one plane are equally inclined to
another plane, they will be equally inclined to the common section
of these planes.

24. PA, PB, PC are three concurrent straight lines, each of
which is at right angles to the other two : PX, Py, PZ are perpen-
diculars drawn from P to BC, CA, AB respectively. Shew that
XYZ is the pedal triangle of the triangle ABC.

25. PA, PB, PC are three concurrent straight lines, each of
which is at right angles to the other two, and from P a perpen-
dicular PO is drawn to the plane of ABC : shew that O is the,
orthocentre of the triangle ABu.

446

EUCLID'S ELEMENTS,

THEOREMS AND EXAMPLES ON BOOK XL

Definitions.

(i) Lines which are drawn on a plane, or through
which a plane may be made to pass, are said to be co-planar.

(ii) The projection of a line on a plane is the locus
of the feet of perpendiculars drawn from all points in the
given line to the plane.

Theobem 1. The projection of a straight line on a plane is itself
a straight line.

Let AB be the given st. line, and XY the given plane.

From P, any point in AB, draw Pp perp. to the plane XY.

It is required to shew that the locus of p is a st. line.

From A and B draw Aa, B6 perp. to the plane XY.

Now since Aa, Pp, B6 are all perp. to the plane XY,

.*. they are par.
And since these par^ all intersect AB,

.*. they are co-planar. xi. 7.

the point p is in the common section of the planes A&, XY ;
that is, j9 is in the st. line ab.

But p is any point in the projection of AB,
.'. the projection of AB is the st. line ah, q.e.d.

XI. 6.

THEOREMS AND EXAMPLES ON BOOK XL

447.

Theorem 2. Draw a perpendicular to each of two straight lines
which are not in the same plane. Prove thaX this perpendicular is the
shortest distance between the two lines.

Let AB and CD be the two straight lines, not in the same plane.

(i) It is required to draw a st. line perp. to each of them.

Through E, any point in AB, draw EF par^ to CD.
Let XY be the plane which passes through AB, EF.

From H, any point in CD, draw HK perp. to the plane XY. xi. IL
And through K, draw KQ parj to tF, cutting AB at Q.

Then KQ is also par^ to CD ; xi.. 9.

and CD, HK, KQ are in one plane. xi. 7.

From Q, draw QP par^ to HK to meet CD at P.
Then shall PQ be perp. to both AB and CD.

For, since HK is perp. to the plane XY, and PQ is par^ to HK,

Constr.
.'. PQ is perp. to the plane XY ; xi. 8.

.*. PQ is perp. to AB, which meets it in that plane, xt. Def 1.
For a similar reason PQ is perp. to QK,
/. PQ is also perp. to CD, which is par^ to QK.

(ii) It is required to shew that PQ is the least of all st. lines
drawn from AB to CD.

Take HE, any other st. line drawn from AB to CD.

Then HE, being oblique to the plane XY, is greater than the
perp. HK. Ex. 1, p. 429.

HE is also greater than PQ. q.e.d.

II.S.E.

448 EUCLID'S ELEMENTS.

Definition. A parallelepiped is a solid figure bounded
by three pairs of parallel plane faces.

Theorem 3. (i) The faces of a parcUlelepiped are parallelograms^
of which those which are opposite are identically equal.

(ii) The four diagonals of a parallelepiped are concurrent and
bisect one another.

Let ABA'B' be a parP^*, of which ABCD, C'D'A'B' are opposite
faces.

(i) Then all the faces ^all he pat^, and the opposite faces shaU
he identically equal.

For since the planes DA', AD' are par^, XT. Def 15.
and the plane DB meets them,
.*. the common sections AB and DC are pari xi. 16.

Similarly AD and BC are pari
.*. the fig. ABCD is a par™,
and AB = DC ; also AD = BC. i. 34.

Similarly each of the faces of the par"^ is a par™ ;
so that the edges AB, C.'D', B'A', DC are equal and par^ :
so also are the edges AD, C'B', D'A', BC ; and likewise AC, BD%
CA', DB'.

Then in the opp. faces ABCD, C'D'A'B',

we have AB = CD' and BC = D'A' ; Proved.

and since AB, BC are respectively par^ to C'D', D'A',

/. the L ABC = the L CD'A'; xi. 10.

.-. the par™ ABCD = the par™ CD'A'B' identically. Ex. 11, p. 70.

(ii) The diagonals AA', BB', CC, DD' shaU he concurrent and
hisect one another.

Join AC and A'C.

THEOREMS AND EXAMPLES ON BOOK XI.

449

Then since AC is equal and par^ to A'C,
.-. the fig. ACA'C is a par" ;
.*. its diagonals AA', CC bisect one another. Ex. 5, p. 70.
That is, AA' passes through O, the middle point of CC.

Similarly if BC and B'C were joined, the fig. BCB'C would be a
par™ ;

.'. the diagonals BB', CC bisect one another.
That is, BB' also passes through O the middle point of CC.

Similarly it may be shewn that DD' passes through, and is
bisected at, O. Q.E.D.

Theorem 4. The straight linefi which join the vertices of a tetra-
hedron to the centroids of the opposite faces are concurrent.

Let ABCD be a tetrahedron, and let g^ f/g, g^, g^ be the centroids
of the faces opposite respectively to A, B, C, D.

Then shall Ag^, Bgg, Cgg, Dg4 he concurrent.

Take X the middle point of the edge CD ;
then gi and g^ must lie respectively in BX and AX,

so that BX = 3 . \g^, Ex. 4, p. 113.

and AX = 3 . Xf/oj
.*. f/igTg is par^ to AB.

And A^i, B(72 must intersect one another, since they are both in
the plane of the A AXB :

let them intersect at the point G.

Then by similar A% AG : Gflri = AB : g^g^

= AX : Xgrg
= 3:1.

.*. Bj72 cuts A.7i at a point G whose distance from <7i = J . Agrj.

Similarly it may be shewn that C73 and D.74 cut kg^ at the same
point ;

.•. these lines are concurrent. Q. e. d.

EUCLID S ELEUENTS.

Ti the dvptkate ratio of their

Let SABCD be a pyramid, and oftcrf the section formed by a
plane drawn par' to the base ABCD.

(i) Then the Jigs. ABCD, al>cd shall be similar.
Because the planes abed, ABCD are paH,

and tlie plane AB6a meets them,

,'. the common sections ab, AB are par'.

Similarly be ia par' to BC ; cd to CD ; and da to DA.

And since ab, be are respectively par' to AB, BC,

.-. the i. a6c = the i. ABC. XI. 10.

Similarly the remaining angles of the fig, aiicd are equal to the
corresponding angles of the fig. ABCD.

And since the A' Sab, SAB are similar,
.■. ab : AB = 8i : SB

= be : BC, for the A"8ftc, SBC are similar.
Or, ah : 6c = AB : BC.

In like manner, be : cd= BC : CD ; and so on.

.'. the figs, ahed, ABCD are equiangular to one another, and
liave their sides about the equal angles proportional ;
.*. they are similar.

(ii) From S draw SxX perp. to the par' planes abed, ABCD
and meeting them at x and X.

Then shall fig. abed -.fin. ABCD = Sx' : SX',

Join ax, AX.

Then it is clear that the A' Sox, SAX are similar.

And the flg. abed : fig, ABCD = ofc= -. AB^ vi. 20,

^aS" i A&,

DEFINITION. 451

Definition. A polyhedron is regular when its faces are
similar and equal regular polygons.

Theorem 6. There cannot he more than five regular polyhedra.

This is proved by examining the number of ways in which it is
possible to foi-m a solid angle out of the plane angles of various
regular polygons ; bearing in mind that three plane angles at least
are required to form a solid angle, and the sum the plane angles
forming a solid angle is less than four right angles. xi. 21.

Suppose the faces of the regular polyhedron to be equHaieral
triangles.

Then since each angle of an equilateral triangle is -3^ of a right
angle, it follows that a solid angle may be formed (i) by three^ (ii) by
foury or (iii) by five such faces ; for the sums of the plane angles

would be respectively (i) two right angles, (ii) f^ of a right angle,

(iii) -g- of a right angle ;

that is, in all three cases the sum of the plane angles would be less

than four right angles.

But it is impossible to form a solid angle of six or more equi-
lateral triangles, for then the sum of the plane angles would be
equal to, or greater than four right angles.

Again, suppose that the faces of the polyhedron are squares.

(iv) Then it is clear that a solid angle could be formed of
three, but not more than three, of such faces.

Lastly, suppose the faces are regular penta^/ons.

(v) Then, since each angle of a regular pentagon is -g^ of a
right angle, it follows that a solid angle may be formed of three such
faces ; but the sum of more than three angles of a regular pentagon
is greater than four right angles.

Further, since each angle of a regular hexagon is equal to ^ of a
right angle, it follows that no solid angle could be formed of such
faces ; for the sum of three angles of a hexagon is equal to four right
angles.

Similarly, no solid angle can be formed of the angles of a polygon
of more sides than six.

Thus there can be no more than^t;e regular polyhedra.

EUCLID S ELEMENTS.

Note on the Regular Polyhedra.

It has /our faces,
fov.r vertioei

It has ei{fA( faces, six

It has Ivtmly faces, t»idi>& verticas, (Atrfsr

! BEGULAR POLYHEDRA.

It has six faces,

It has tweive (acea, Iwenly vertices, (AiV(y edges.

454 EUCLID'S ELEMENTS.

Theoiiem 7. //* F denote the nurtiber of faces j E of edges, and V
of vertices in any polyhedron, then wUl

E + 2=F+V.

Suppose the polyhedron to be formed by fitting together the faces
in succession : suppose also that Er denotes the number of edges, and
Vr of vertices, when r faces have been placed in position, and that
the polyhedron has n faces when complete.

Now when one face is taken there are as many vertices as edges,
that is, Ei = Vi.

The second face on being adjusted has two vertices and one edge
in common with the first ; therefore by adding the second face we
increase the number of edges by one more than the 'number of
vertices; ,.. E^-\J^^i,

Again, the third face on adjustment has three vertices and two
edges in common with the former two faces ; therefore on adding the
third face we once more increase the number of edges by one more
than the number of vertices ;

.-. E3-V3=2.

Similarly, when all the faces but one have been placed in position,

E«-i-V„_i=w-2.

But in fitting on the last face we add no new edges nor vertices ;

.*. E = E„_i, V = V„_i, and F=w.

So that E-V = F-2,
or, E + 2 = F+V.
This is known as Evler^s Theorem.

Miscellaneous Examples on Solid Geometry.

1. The projections of parallel straight lines on any plane are
parallel.

2. If ah and cd are the projections of two parallel straight lines
AB, CD on any plane, shew that AB : CD = a6 : cd.

3. Draw two parallel planes one through each of two straight
lines which do not intersect and are not parallel.

4. If two straight lines do not intersect and are not parallel, on
what planes will their projections be parallel ?

5. Find the locus of the middle point of a straight line of
constant length whose extremities lie one on each of two non-inter-
sectiag straight lines.

MISCELLANEOUS EXAMPLES ON SOLID GEOMETRY. '455

6. Three points A, B, C are taken one on each of the conter-
minous edges of a cube : prove that the angles of the triangle ABC
are all acute.

7. If a parallelepiped is cut by a plane which intersects two
pairs of opposite faces, the common sections form a parallelogram.

8. The square on the diagonal of a rectangular parallelepiped is
equal to the sum of the squares on the three edges conterminous
with the diagonal.

9. The square on the diagonal of a cube is three* times the square
on one of its edges.

10. The sum of the squares on the four diagonals of a parallele-
piped is equal to the sum of the squares on the twelve edges.

11. If a perpendicular is drawn from a vertex of a regular
tetrahedron on its base, shew that the foot of the pei*pendicular will
divide each median of the base in the ratio 2:1.

12. Prove that the perpendicular from the vertex of a regular
tetrahedron upon the opposite face is three times that dropped from
its foot upon any of the other faces.

13. If A P is the perpendicular drawn from the vertex of a regular
tetrahedron upon the opposite face, shew that

3AP2=2a2,

where a is the length of an edge of the tetrahedron.

14. The straight lines which join the middle points of opposite
edges of a tetrahedron are concurrent.

15. If a tetrahedron is cut by any plane parallel to two opposite
edges, the section will be a parallelogram.

16. Prove that the shortest distance between two opposite edges
of a regular tetrahedron is one half of the diagonal of the square on
an edge.

17. In a tetrahedron if two pairs of opposite edges are at right
angles, then the third pair will also be at right angles.

18. In a tetrahedron whose opposite edges are at right angles in
pairs, the four perpendiculars drawn from the vertices to the opposite
faces and the three shortest distances between opposite edges are
concurrent.

19. In a tetrahedron whose opposite edges are at right angles,
the sum of the squares on each pair of opposite edges is the same.

20. The sum of the squares on the edges of any tetrahedron is
four times the sum of the squares on the straight lines which join the
middle points of opposite edges.

456 Euclid's elements.

21. In any tetrahedron the plane which bisects a dihedral angle
divides the opposite edge into segments which are proportional to
the areas of the faces meeting at that edge.

22. If the angles at one vertex of a tetrahedron are all right
angles, and the opposite face is equilateral, shew that the sum of the
perpendiculars dropped from any point in this face upon the other
three faces is constant.

23. Shew that the polygons formed by cutting a prism by parallel
planes are equaL

24. Three straight lines in space OA, OB, OC, are mutually at
right angles, and their lengths are a, h, c: express the area of the
triangle ABC in its simplest form.

25. Find the diagonal of a regular octahedron in terms of one of
its edges.

26. Shew how to cut a cube by a plane so that the lines of
section may form a regular hexagon.

27. Shew that every section of a sphere by a plane is a circle.

28. Find in terms of the length of an edge the radius of a sphere
inscribed in a regular tetrahedron.

29. Find the locus of points in a given plane at which a straight
line of fixed length and position subtends a right angle.

30. A fixed point O is joined to any point P in a given plane
which does not contain O ; on OP a point Q is taken such that th&
rectangle OP, OQ is constant : shew that Q lies on a fixed sphere.

OLASQOW : PRINTED AT THE UNIVERSITY PRESS BY ROBERT MACLEHOSK AND CX?. LTDU

WORKS BY H. S. HALL, M.A., and P. H. STEVENS, M.A.

A SCHOOL Geometry, based on the recommendations of
the Mathematical Association, and on the recent report of
the Cambridge Syndicate on Geometry.

Parts I. and H. — Part I. Lines and Angles. Rectilineal Figures.
Part II. Areas of Rectilineal Figures. Containing the sub-
stance of Euclid Book L Crown Svo. Is. Qd.

Part HI. — Circles. Containing the substance of Euclid Book IH.
. 1-34, and part of Book IV. 1«.

Parts L, II., III. in one volume. 25. 6ti.

Part IV. — Squares and Rectangles. Geometrical equivalents of
Certain Algebraical Formulae. Containing the substance of
Euclid Book II., and Book HI. 35-37. 6d.

Parts III. and IV. in one volume, la. 6d.

Parts I. -IV. in one volume. 35.

Part V. — Containing the substance of Euclid Book VI. 1«. 6d.

Parts IV. and V. in one volume. 25.

Parts I.-V, in one volume. 4«. 6d.

Part VI. — Containing the substance of Euclid Book XL 1-21, to-
gether with Theorems relating to the Surfaces and Volumes
of the simpler Solid Fighres. 25.

A TEXT-BOOK OF EUCLID'S ELEMENTS, including
Alternative Proofs, together with Additional Theorems
and Exercises, classified and arranged. Third Edition
enlarged. Books I. — VI. XI. and XII. Props. 1 and 3.
Globe Svo. 45. Qd.
Also in parts, separately, as follows : —

Book I. - - l5.
Books I. and II. l5. 6(f.
Books II. and III. 25.
Books L— III. - 25. 6fl?.
Books I.— IV. - 35.
Sewed, 25. 6c?.

Books III. and IV. - 25.
Books III.— VI. - 35.
Books IV.— VI. - 25. Qd.
Books v., VI., XL and XIL
1 and 3. - - 25. Qd.
Book XI. - - l5.

JOURNAL OF EDUCATION— '"HhQ most complete introduction to plane
geometry based on Euclid's Elements that we have yet seen."

A KEY TO THE EXERCISES AND EXAMPLES CON-
TAINED IN A TEXT-BOOK OF EUCLID'S ELE-
MENTS. Books I.— VI. and XL By H. S. Hall, M.A.,
and F. H. Stevens, M.A., Masters of the Military Side,
Clifton College, down Svo. 85. 6d Books L— IV. 6*. 6fl?.
Books VI. and XL 35. 6d

AN ELEMENTARY COURSE OF MATHEMATICS. Com-
prising Arithmetic, Algebra, and Euclid. By H. S. HatJj,
M.A., and F. H. Stevens, M.A. Globe Svo. 2*. %d.

MACMILLAN AND CO., Ltd., LONDON,

WORKS BY H. S. HALL, M.A., and S. R. KNIGHT, B.A.

EIGHTH EDITION, Revised and Enlarged.

ELEMENTARY ALGEBRA FOR SCHOOLS (containing a
chapter on Graphs). Globe 8vo. (bound in maroon -coloured
clotn), 3«. 6d. With Answers (bound in greeu-coloured
cloth, 4«. 6d,

NA yiTBJ?.--" This is, in our opinion, the best Elementary Algebra for school use.
It is the combined work of two teachers who have had considerable experience
of actual school teaching . . . and so successfully grapides with difficulties which
our present text*books in use, from their authors Letcking such experience, ignore
or slightly touch upon. . . . We confidently recommend it to mathematical teasers,
who, we feel sure, will find it the best book of its kind for teaching purposes."

ANSWERS TO EXAMPLES in ELEMENl'ARY ALGEBRA
Fcap. 8va Sewed Is,

SOLUTIONS OF THE EXAMPLES IN ELEMENTARY
ALGEBRA FOR SCHOOLS. Crown 8vo. Ss. ed,

HIGHER ALGEBRA. A Sequel to Elementary Algebra for
Schools. Fifth Edition, revised and enlarged. Crown 8va

ATHENJEUM.—*' The Elementary Algebra, by the same authors, is a work of
su(di exceptional merit that those acquainted with it will form hign expectations
of the sequel to it now Issued. Nor will they be disappointed. Of the authors'
Higher Algebra as of their Elementary Alfiebru, wo unhesitatingly assert that it is
by far the best work of its kind with which we are acquainted. It supplies a
waint much felt by teachers. "

SOLUTIONS OF THE EXAMPLES IN HIGHER AL-
GEBRA. Crown 8vo. 10s. (id,

ALGEBRA FOR BEGINNERS. Globe 8vo, 28. With Answers.
25. 6d.

SCHOOLMASTER.— "To teachers who have had experience of either the
Elementary or the Higher Algebi-a it will only be necessary to say that this book
is marked by the same qualities which have brought tiiese works into such
deserved repute. To those who are still in ignorance of these books, we can say
that for dear, simple, and concise explanation, convenient order of subject*
matter and copious and well-graduated exercises, these books have, to say
the least, no superiors.

ABERDEEN FREE PRESS.—" To give the learner a sufficient knowledge of the
ordinary algebraical processes without presenting him with too many of their
difficulties requires a thorough knowledge of what is necessary for a banner,
and of what may be omitted until a more advanced stage. A very slight exami*
nation of the book shows that the authors poRsess this qualification in an
eminent degree. . . . The definitions and explanations of symbols are full and
simple. . . . An important feature of the work is the great number of problems
given to show the practical applications of the science."

MACMILLAN AND CO., Ltd., LONDON.

WORKS BY H. S. HALL, M.A., and S. R. KNIGHT, B. A.

ALGEBRAICAL EXERCISES AND EXAMINATION
PAPERS. With or without Answers. Third Edition,
revised and enlarged. Globe 8vo. 28. 6d.

IRISH TEACHERS^ JOURNAL.—*' We know of no better work to place in the
hands of junior teachers, monitors, and senior pupils. Any person who works
carefully and steadily through this book could not possibly fail in an examination
of Elementary Algebra. . . . We congratulate the authors on the skill dis-
played in the selections of examples."

SCHOOLMASTER— "We can strongly recommend the volume to teachers
seeking a well-arranged series of tests in Algebra."

ARITHMETICAL EXERCISES AND EXAMINATION
PAPERS. With an Appendix containing Questions in
LOGARITHMS AND MENSURATION. With or without
Answers. Third Edition, revised and enlarged. Globe 8vo.

28. ed.

CAMBRIDGE REVIEW.— " AM the mathematical work these gentlemen have
given to the public is of genuine worth, and these exercises are no exception to
the rule. The addition of the logarithm and mensuration questions add greatly
to the value."

ELEMENTARY TRIGONOMETRY. Third Edition. Globe
8vo. 4s. 6d.

ACADEMY.— *' It is drawn up with that skill which shows the writers to
be past masters in the art of teaching, and is quite on as high a platform
for Tiigonometry as their previous books on Algelnra are for that subject. We
have read the text with care and have thoroughly enjoyed it."

EDUCATIONAL REVIEW—" On the whole it is the best elementary treatise on
Trigonometry we have seen."

SOLUTIONS OF THE EXAMPLES IN ELEMENTARY
TRIGONOMETRY. Crown 8vo. 8s. 6d.

ALGEBRAICAL EXAMPLES Supplementary tp Hall and
Knight's ALGEBRA FOR BEGINNERS and ELEMEN-
TARY ALGEBRA (Chaps. L-XXVII.). By H. S. Hall,
M.A. "With or without Answers. Globe 8vo. 2s,

A SHORT INTRODUCTION TO GRAPHICAL ALGEBRA.
By H. S. Hall, M.A. Third Edition revised and enlarged.
Globe 8vo. Is. [^^» *^ i^ Press,

MACMILLAN AND CO., Ltd., LONDON.

WORK BY H. S. HALL, M.A., and R. J. WOOD, B.A.

ALGEBRA FOR ELEMENTARY SCHOOLS. Globe 8vo.
Parts 1., II., and III., 6d. each. Cloth, Sd, each. Answers.
4d, each.

BEAD TEACHER, — "Any boy working conscientiously through this book will
have laid a solid foundation for more advanced effort."

TEACHERS* AID. — "The aim of the authors is to provide an easy and well-
graduated course of Algebra suited to the requirements and capacity of those who
take it as a specific subject in elementary schools. The general plan and arrange*
ment of Hall & Knight's well-known elementary text-book have been closely
followed with such changes as the special needs of elementary pupils demand. It
is needless to say that the book in every way is a creditable production."

MADRAS EDUCA TTONAL NEWS.—" The examples are many and well chosen.
. It is well calculated to serve the purpose for which it is intended."

WORKS BY F. H. STEVENS, M.A.

AiENSURATION FOR BEGINNERS, with the Rudiments
of Geometrical Drawing. Globe 8vo. Is, 6d,

THE EDUCATIONAL TIMES (March, 1897).— "A considerable amount of
ground is covered, and the whole is written with rare judgment and clearness.
We should judge that it would be practically impossible to produce a book better
adapted to its purpose."

THE GUARDIAN (January 18th, 1897).— "The little book is really excellent
Mr. Stevens seems to us to have chosen just the elements that are of use in
every>day life, and, in the notes and examples he has worked out, to have given
sufficient illustraraon of right methods. Exercises are numerous and dis-
creetly tabulated ; and — a point worth noticing — a good many are set f(Mr graphic
solution."

UNIVERSITY CORRESPONDENT.—" The main book seems to us to be admir-
ably written and got up in attractive form, with just such differences of type as
are a real help."

EDUCATIONAL NEWS.— *' This little treatise will be found thoroughly prac-
tical . . . the explanations are clear and to the point, and the examples
numerous and graded. It is worth the attention of teachers and pupil-teachers,
who have this subject to take. The book is in i^ respects worthy ofcommenda-
tion."

ELEMENTARY MENSURATION. Globe 8vo. Zs, ed.

THE OXFORD MAGAZINE (May 27th, 1896).— "Mr. Stevens seems bent on
making students understand the meaning of the processes employed ; and his
examples are excellent. This is by far the best book extant on the subject."

SCHOOL GUARDIAN.—*' The volume is systematically arranged, clearly and
concisely written, and fully illustrated by numerous diagrams ana a variety of
examples and exercises. It is in every way a creditable piece of work, and
worthy of commendation."

MACMILLAN AND CO., Ltd., LONDON.

i^ «1 .Ha4 1904 n

A tttct-book of Budfcfi •!■«-« C.l

JtMloni

f^Hartm

3 6105 030 435 510

DATE DUE

STANFORD UNIVERSITY LIBRARIES
STANFORD, CALIFORNIA 94305-6004

Mbdufi

SSSSEBSS999RS.

Internet Archive Audio

Featured

Top

Images

Featured

Top

Software

Featured

Top

Texts

Featured

Top

Video

Featured

Top

Mobile Apps

Browser Extensions

Archive-It Subscription

Save Page Now

Full text of "A Text-book of Euclid's Elements for the Use of Schools"

See other formats